RADIOACTIVITY Some elements are unstable. They spontaneously emit particles like alpha, beta or positron and change to other elements. This phenomenon is called radioactivity. We know from the chapter Periodic Table that all the isotopes of all the elements above atomic number 83(Bi) are radioactive and are unstable. From 84-92, the elements are naturally radioactive and above 92 are synthetically radioactive. However some isotopes of elements below atomic number 83(Bi) are also radioactive. For example, C12 and C13 isotopes are stable while C14 isotope is radioactive; H1 and H2(deuterium) are stable while H3(tritium) isotope is radioactive; K39 isotope is stable while K40 isotope is radioactive and so on. Majority of the lighter isotopes are stable and are not radioactive. A few of them are unstable. But above Bi, i.e from Po(84) onwards, no isotope is stable. The isotopes are also called nuclides and so in this chapter we shall use the term nuclide in place of isotope very often. The particles alpha, beta etc. emitted from the radioactive nuclide come out of the nucleus of the nuclide. Alpha and beta are particles, not true rays which look like rays because the partcles travel with vey high speed. Gamma is a electromagnetic radiaon i.e it is a true ray. Let us know first about the different radioactive emissions.

ALPHA(α α) PARTICLES: An alpha particle is a particle consisting of 2 protons and 2 neutrons which come out of the nucleus with high speed. The alpha particle is given a symbol 2He4 because it resembles with dipositive He++ ion. A He++ ion contains 2 protons and 2 neutrons (i.e 2 electrons removed from a He atom) like an alpha particle, hence the symbol. When alpha particle is lost by the nuclide, the new nuclide called the daughter nuclide contains two protons less and so the new element is placed two places left of the parent element in the periodic table. 92

U238 ----------> 2He4(alpha particle) + 90Th234

Note that the subscript gives the atomic number i.e the proton number while the superscript gives the mass number(proton number + neutron number). When Uranium-238 isotope emits one alpha particle, the daughter nuclide contains 90 protons and the new element is Th-234 isotope. Note that in nuclear reactions of the type shown above, the reader should see that the mass number in the LHS should be same as the total mass number in the RHS(i.e the total of superscripts of two sides should be equal), similarly the proton numbers of LHS should be same as the total proton numbers of RHS(i.e the total of subscripts of two sides should be same). 238 = 4 + 234(mass number) and 92 = 2 + 90(proton number) One thing is to be borne in mind that when one alpha particle is emitted by the nucleus of the atom, the total positive charge of the atom decreases by two. But the number of electrons remains the same(92 in the above case) and now electron number is more than the proton number by two. So what happens to the electrical neutrality of the daughter nuclide? Does it become dinegative charged ion? No, the electrical neutrality is maintained soon after the emission of one alpha particle. The excess electrons are removed from the valence shell of the daughter nuclide to the surrounding atmosphere to equalise the number of protons with number of electrons(90 in this case). Hence the daughter nuclide becomes a neutral atom like the parent element and not an anion.

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More Examples: Pu239 --------> ? + 2He4(alpha) ( ? = 92U235) 94 When Plutonium 239 isotope emits an alpha particle, the daughter nuclide is Uranium 235 isotope which is different from U-238 isotope taken in the first example. Note that U-238 isotope is an alpha emitter(1st example) and Pu-239 isotope(2nd example) is also an alpha emitter. Those nuclides which emit alpha particles are called alpha emitters. They cannot be beta emitters, i.e those isotopes cannot emit beta particles. Another important thing you remember that alpha particle is emitted only by heavier nuclides i.e usually nuclides above atomic number 82(Pb), and not by any ligher nuclide such as 6C14 or 19K40 etc. SAQ 1: The following nuclides are alpha emitters. Predict the daughter nuclides and write the nuclear equation when each nuclide undergoes alpha decay. (i)90Th230 (ii)90Th232 (iii)88Ra226 (iv) 86Rn220

BETA(β β) PARTICLES: A beta particle is identical with an electron. It has negligible mass(nearly 0) and charge equal to -1. It is given the symbol -1e0. Always the subscript represents the charge and superscript represents the mass. The question is how a beta particle which is nothing but an electron is produced in the nucleus. A beta particle is not the electron already present in the atom in the extra nuclear part. It is a new particle looking like an electron in mass and charge produced at the nucleus and is expelled out from it at a tremendous speed. They are also wrongly called β-rays. A neutron in the nucleus is transformed to a proton and an electron(beta particle). 0

n1 --------> +1p1 + -1e0

(beta particle)

Thus when this beta particle is driven out of the nucleus, the nucleus now contains one proton excess than the parent nuclide, although the mass remains nearly same. So the daughter element contains one excess proton and occupy one place right in the periodic table. Look to this example. 90

Th234 --------> 91Pa

234

+ -1e0 (beta particle)

When Th-234 isotope emits a beta particle, the daughter nuclide occupies one place right of it, that is Pa-234(protactinium) isotope is formed. Note that the parent nuclide had 90 electrons in the extranuclear part(90Th232) but now the daughter has 91 protons with same number(90) of electrons. What will happen to the electrical neutrality of the daughter nuclide? It becomes electrically neutral by absorbing one electron from surrounding atmosphere and equalising the number of protons with number of electrons. Remember that atmosphere is a great reservoir of electrons. It can donate and accept electrons whenever required. More example: C14 --------> ? + -1e0 (beta) (? = 7N14) 6 Po218 -------> ? + -1e0 (beta) (? = 83Bi218) 82 Note that the nuclide which is a beta emitter like C-14 isotope and Po-218 isotope are only

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beta emitters. They can never be alpha emitters. Unlike alpha emission, which takes place only by heavier nuclide(>Pb), beta emission takes place by both lighter nuclide( eg. C-14) and heavier nuclide(Po-218). SAQ 2:The following nuclides are beta emitters. Write the nuclear equations and predict the daughter nuclides. (i) 90Th234 (ii) 19K40 (iii)91Pa234 (iv)27Co60 212 (v) 83Bi SAQ 3: 92U238 isotope is a ______emitter and its daughter nuclide after alpha emission is a _____emitter. Is there any relationship between the nature of emission of the parent and daughter nuclides?

POSITRON EMISSION: Positron is another particle which is emitted from certain radioactive nuclides. Positron is not a fundamental particle like electron or proton. It is only produced during radioactive decay of certain radioactive nuclides. Positron is analogous to an electron in mass(negligible mass) but also analogous to a proton in charge(+1 charge). It is given the symbol +1e0. Now the question is how a positron is produced in the nucleus. A proton is transformed to a neutron and a positron. Look to the following equation. p1 --------> 0n1 + +1e0 (positron) +1 After the positron is driven out, the daughter nuclide has proton number less than that of the parent nuclide by one, although mass remains nearly constant. Thus the daughter nuclide is a new element occupying one place to the left of the periodic table. Let us look to this example. Na22 --------> +1e0 + 10Ne22 11 Al25----------> +1e0 + 12Mg25 13 Note that Na-22 and Al-25 isotopes are positron emitters? The normal Na and Al that we know are Na-23 and Al-27 isotopes which are stable and not radioactive. But Na-22 and Al-25 isotopes which are prepared artificially are radioactive and are positron emitters. When they expel a positron from their nuclei, they are converted to new elements Ne-22 and Mg-25 isotopes respectively. So in general when a positron is emitted the new daughter element occupies one place left of it in the periodic table. After the loss of a positron, the nucleus now has one proton short compared to the parent but the number of electrons remains the same. The electrical neutrality of the daughter species is maintained by the immediate loss of one electron from the valence shell of the daughter to the surrounding atmosphere so that the number of protons and electrons remain the same in the daughter nuclide. Thus the daughter is electrically neutral and not a negative charged ion. Note that the nuclide which emits positron is a positron emitter. It cannot be either alpha or the beta emitter. Only lighter nuclides( 28Ni60* (1st excited state) + 28 60 60 Ni * --------> 28Ni (ground state) + γ photon 28 Thus after the emission of a beta particle by a Co-27 isotope, the new nuclide Ni-60 climbs up to the 2nd excited state. It comes back to ground state in two successive jumps and thereby emit two photons of gamma rays. To conclude, gamma is not a particle like alpha, beta and positron. It is an energy radiation which is released after the emission of alpha, beta or positron. An alpha emitter can emit gamma ray photon, a beta emitter can emit a gamma ray photon. Note that, not all radioactive nuclides emit gamma ray photon after the emission of particles like alpha, beta. There are some which do so, not all. For example, C-14 isotope does not emit a gamma photon after emitting a beta particle, while Na-24(artificial) isotope emits a gamma photon after emitting a beta particle. (2) The other way by which a gamma photon is formed is positron-electron annihilation. This happens in case of positron emitters in isotopes having lower n/p ratio than expected. In some cases, positron instead of going out of the atom unites with an electron of the valence shell of an atom. When positron unites with an electron, both of them are destroyed in a process called annihilation and the mass is coverted to energy. The positron-electron annihiliation produces two gamma photons moving in opposite directions. 27

SAQ 4: Write the radioactive decay reaction for the following nuclide. The particle emitted in each case has been mentioned within bracket. Refer a periodic table to locate the position of the daughter nuclide. (i)12Mg27(β) (ii)83Bi210(β) (iii)84Po213(α) (iv) 8O15(positron) 200 82 (v)79Au (β) (vi)35Br (β) (vii)88Ra226(α) (viii)84Po210(α) 38 122 (ix)19K (positron) (x)53I (positron) (xi)26Fe60(β), (xii)93Np230(α) (xiii)23V46(positron) (xiv)87Fr221(α)

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PROPERTIES OF ALPHA, BETA, POSITRON AND GAMMA RAYS: ALPHA: These are like He++ ions consisting of two protons and two neutrons. They are ejected out from the nucleus with high velocities in the range of 1.7 X 107 cm/sec which is 6% of the velocity of light. They are deflected to the negative plate of an electric field. The deviation towards the negative potential is smaller compared to beta particles as they possess higher kinetic energy. They can penetrate through thin metal sheets. Do you remember the Rutherford's gold foil experiment? Alpha particles penetrated through a thin gold foil of 0.00004cm thickness. These have the least penetrating power and can be stopped by aluminium sheet of 0.1mm thick. It cannot penetrate metal sheets above this thickness. Alpha particles cause ionisation of the gas through which they pass. Its ionising power is the maximum because it contains the highest kinetic energy. They affect the photographic plate. When alpha particles strike on a ZnS screen, produces luminescence(ZnS glows). BETA: These are nothing but electrons( -1e0) having -1 charge and negligible mass. They travel with high velocity often higher than alpha particles(nearly equal to velocity of light- 99% of velocity of light). They are deflected to the positive plate of an electric field and the extent of deviation is more as they posses less kinetic energy than alpha particles. They have more penetrating power than alpha particles. They can penetrate through thicker metal sheets than alpha particles.They can be stopped by aluminium sheet of 1cm thick. Since they possess lower kinetic energy(due to negligible mass, E = mc2), they ionise gases to a smaller extent than alpha particles. They also affect photographic plate to a greater extent and also produce glow in ZnS screen but to a lesser extent than alpha particles. POSITRON: These carry +1 charge like protons and negligible mass like electrons and are represented as e0.Therefore they are sometimes called as positive electrons. They travel with high speed like +1 beta particles and are deflected towards the negative plate of the electric field. Other properties are analogous to beta particles.

GAMMA RAYS: These are not particles and are high energy(low wavelength) electromagnetic radiations. They do not carry any charge or mass. They travel with the velocity of light. They are not deflected by electric or magnetic field. They have the highest penetrating power, they can pass through much thicker aluminium sheets. They can be stopped by aluminium sheet of 100cm thickness. They ionise the gases but to the smallest extent as they do not carry any mass. They affect to the smallest extent to the photographic plate and also produce glow in ZnS screen.

TYPES OF RADIOACTIVITY: There are two types of radioactivity, i.e (i)Natural radioactivity: Those nuclides which are present in the nature and are radioactive are called naturally radioactive. The phenomenon is called natural radioactivity. All the isotopes of all elements beyond atomic number 83(Bi) are naturally radioactive. However some lighter isotopes such as 6C14, 19K40, 43Tc99 etc. are also naturally radioactive. (ii) Artificial Radioactivity: Those radioactive nuclides which are artifically prepared give rise to the phenomenon of artificial radioactivity. Majority of the lighter nuclides that are

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radioactive such as 5B10, 11Na22, 27Co60 etc. are all artificially prepared and are radioactive.

THEORY OF RADIOACTIVITY: It is very surprising to believe that although protons are positively charged particles, remain together in a small space called nucleus without repelling each other. What factor is responsible for binding the protons together? It is the neutral particles- neutrons which remain in between protons and are responsible to bind the protons in the nucleus. Neutrons and protons are jointly called nucleons. The neutrons act like cement to bind the protons together, so that we get a stable nucleus. For very lighter nuclides i.e upto atomic number 20(Ca), when the proton number is small and the repulsion between the protons is not high, the number of neutrons necessary to make stable nuclei is nearly equal to the number of protons present. For example, C12, 8O16, 20Ca40 etc. in which the number of protons is equal to the number of neutrons. 6 Although many exceptions are there like 5B11, 11Na23, 13Al27 etc. in which the number of neutrons is greater than number of protons by one, the nuclides are still stable. Such deviations are not considered abnormal, as the law of nature at times does not follow a fixed order or rule as predicted by human beings and it is not possible to explain its cause. Anyway, upto atomic number of 20(Ca), the neutron to proton ratio(n/p) is nearly equal to one. As the atomic number increases beyond 20, the repulsion between the protons become more and more severe. To overcome that more and more number of neutrons are necessary to make stable nuclides. So the neutron to proton ratio(n/p) steadily increases for the stable nuclides as we move on to higher atomic numbers. Let us take the examples of 26Fe56, 35Br79, 80Hg200 which are stable nuclides. The number of neutrons in them respectively are 30, 44 and 120. If we take the n/p ratio of these three nuclides, they are 30/26=1.153 for Fe, 44/35=1.257 for Br and 120/80=1.5 for Hg. So, what do we find here? For getting stable nuclide, the neutron number should be more and more i.e the n/p ratio has to increase gradually as the proton number increases beyond 20. It should not remain close to 1. This is required to overcome the repulsive forces between protons which become more and more severe. Speaking in a different way, we can say that Fe nucleus would have been unstable if n/p ratio would have been 1 instead of 1.153 i.e if it would have 26 neutrons in stead of 30. Similarly Br and Hg nuclei would have been unstable if they would have neutrons less than 44 and 120 respectively. Upto atomic number 83(Bi), we find at least one isotope of every element(except Tc and Pm) which are stable. The n/p ratio gradually increases from nearly 1 at Ca(20) to nearly 1.5 at Bi(83). But look to the law of nature after atomic number 83!!! For elements beyond Bi(83), the repulsive forces become so severe that, no excess number of neutrons is able to bind the protons and make a stable nucleus. So all the isotopes of all the elements beyond Bi(83) are unstable nuclides and are spontaneously radioactive. They emit alpha or beta particles and change to another element. The whole purpose behind this emission and transformation is to acquire more and more stability in the daughter nuclei. The emission of radioactive particles (a and β) continues in sequence by a parent to give daughter, then daughter giving grand daughter and grand daughter giving great grand daughter and so on until the last nuclide acquires atomic number 83 or less. Usually the nuclides ultimately transform to one of the isotopes of lead(Pb-82). Look to this sequence of natural radioactivity. 92

U238 -----α ------> 90Th234 ------β------>91Pa234 ------β-----> 92U234 ------α---->90Th230

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------α-----> 88 Ra 226 ------α-------> 86 Rn 222 ------α------> 84 Po 218 ------β----> 85 At 218 ------α----->83Bi214 -----α------> 81Tl210-----β----->82Pb 210-----β------>83Bi210----α----> 81Tl206 ------β----> 82Pb206(Stable) U238 isotope is an alpha emitter and its daughter 90Th234 is a beta emitter and its daughter 91Pa234 is also a beta emitter. In this way, every daughter produced from its parent is unstable and decays either by emitting an alpha or a beta particle. At last when a stable isotope is formed, radioactivity is stopped. 82Pb206 isotope is the end nuclide in the above series. This is called Uranium(238) series. The stability of different nuclides is best known from the graph of neutron versus proton for different nuclides as shown in the following diagram called the Band of Stablity. The 450 straight line is the n/p=1 line, that means if any nuclide will fall on this line, its neutron and proton numbers are equal. Upto proton number 20(Ca), n/p is nearly 1, and nuclides fall more or less near the 450 line but beyond Ca the n/p ratio gradually increases as explained before. The dotted band in the graph is called the band of stability. Because the n/p ratio of stable nuclides above Ca increases steadily and so the stability zone, instead of going straight, rises up in a curved path whose slope gradually increases after calcium. Note that, in stead of a curved line, there is curved band as always there are exceptions and deviations to the general rule that has been theoretically formulated. Any nuclide which falls inside the stability band is a stable nuclide and not radioactive. The stability band is shown upto atomic number 83. That means upto atomic number 83, we get stable nuclides, although some nuclides within this region too are unstable. But above 83(Bi), there is no isotope of any element which is stable and hence all are radioactive. Some of them emit alpha and some beta particles and the emission successively continues until the last nuclide falls within the stability band( -1e0 + 13Al27 12 Mg-27 isotope is unstable and radioactive as it has more number of neutrons than expected of its stable nuclide(12Mg24). It is transformed to the stable 13Al27 isotope by emitting a beta particle. So remember that any nuclide whose mass number appears to be more than what is commonly known is radioactive and a beta emitter. Look to another example. C14 ---------> -1e0 + 7N14 6 C-14 lies above the band of stability and hence is radioactive. It decays by beta emission, so that it is converted to a stable N-14 nuclide. Solve the following SAQs.

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SAQ 6: Why the following are beta emitters? Give the daughter nuclides formed in each case. H3, 35Br82, 26Fe60, 27Co60, 47Ag111 1 Case-II: Unstable nuclides which lie below the band of stability(positron emission): The nuclides which lie below the band of stability(n/p ratio less than expected) i.e mass numbers are less than those of their stable nuclides, are positron emitters. By emitting a positron, it tends to stabilise itself. You know that a positron particle is formed by the conversion of a proton into a neutron and a positron particle. After the loss of positron, the nucleus gains one neutron and loses a proton. Thus n/p ratio increases and the daughter nuclide tends to become more stable. Look to this example. 8

O15 ---------> +1e0(positron) + 7N15,

19

K38 -------->

e0 + 18Ar38

+1

In these cases the n/p ratio of each daughter element is more than that of its parent nuclide. Calculate for yourself to verify this. Both the daughter nuclides formed in the above examples are more stable than their parents. SAQ 7: Explain why the following are positron emitters? Give the daughter nuclide formed in each case. P28, 7N12, 10Ne18, 11Na21, 21Sc41, 16S30, 23V46, 25Mn50 15 SAQ 8: Which of the following is a beta emitter and which a positron emitter and why? C14, 5B10, 11Na22, 11Na24, 19K40, 19K37, 12Mg27, 25Mn56, 53I129, 15P28, ,21Sc41, 27Co60 6 The stable nuclides have the mass numbers as follows: C-12, B-11, Na-23, K-39, Mg-24, Mn-55, I-127, P-31, Sc-45, Co-59,

HALF LIFE PERIOD All radioactive elements decay in a very interesting style. The process of their disintegrtion is called a first order reaction. The nature of a first order reaction is that the time required to complete half of the reaction is always same. For example, let us take 100g[A0]m of a radioactive substance (say 92U238). Let the time taken for the disintegration of half of it is T1 years. So after T1 years, we find that 50gms have disintegrated to its daughter and remaining 50gms of U[A0/2] would be present. Now guess what time is required to complete half of 50gms which is present now. It will take another T1 years. That means for disintegrating half of 50gms i.e 25 gms it will take the same time(T1 years) as it took for disintegrating 100gms to 50gms. So after two half life periods i.e 2T1 years, the amount of 92U23 left is 25gms[A0/4]. Now what time is required for disintegrating half of this value? Another T1 years are required for disintegrating half of 25gms i.e 12.5 gms and leave back 12.5gms of the parent(U238)nuclide. So after 3 half life periods(3T 1), the amount left will be

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12.5gms[A0/8] .This trend continues until all the substance is exhausted. Truly speaking, it will take infinite time for the completion of the reaction, i.e to totally exhaust the parent nuclide, but practically the reaction is nearly 99.9% complete after 10 half life periods. This type of graph is called exponential decay graph. The disintegration of radioactive elements and the concept of half life periods are depicted in the given diagram. After nearly 5 half life periods, the graph comes close to x-axis which indicates that the process is nearly going to be completed. Very interestingly the half life periods of radioactive elements vary from few seconds to several years. It varies from radioactive nuclide to radioactive nuclide. The half life periods of some common radioactive nuclides are given below. Half Life periods of Some Naturally Radioactive Elements: NUCLIDE Uranium(U-238) Radium(Ra-226) Radium(Ra-228) Thorium(Th-230) Radon(Rn-222) Carbon(C-14)

HALF-LIFE 4.5 X 109 yrs 1590yrs 6.7yrs 8000yrs 3.82days 5,730 yrs

EMISSIONS alpha alpha beta alpha alpha beta

Half Life periods of Some Artificially Radioactive elements NUCLIDE Tritium(H-3) Oxygen (O-15) Sodium(Na-24) Phosphorus(P-32) Iodine(I-131)

HALF-LIFE 12.26 yrs 123 secs 14.96 hrs 14.28 days 8 days

EMISSIONS Beta Positron Beta Beta Beta

We find that for 92U238 isotope the half life is so high i.e 4.5 X 109 years, while for 86Rn222 it is few days(3.82 days ) while for 8O15 it is 123secs. The element which has a high half life will take a long time to exhaust while those having small half life will end soon. That is why many naturally radioactive elements (Tc, Pm) which have very short half life periods are almost absent now. Such elements have almost completely died due to their small half lives. There is a short and handy formula which you can use at this stage to find out the amount of a radioactive element left after a particular time period of observation. For that you have to know the half life period of the element.

A t = A0 1 2

t T0 .5

Where At = amount of substance left, A0=amount of substance initially taken, t= timeof observation, T0.5= half life period , t/T0.5 = number of half lives

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Example: O-15 is a positron emitter having half life of 123secs. How much of O-15 will remain after 10minutes and 15 secs if we take initially 2gms of it? Solution: Time =t = 10X60 + 15 = 615 secs. At = 2(1/2)615/123 = 2(1/2)5 = 2/32 = 1/16gm. So the amount left after 10minutes 15 secs will be 1/16gm. SAQ 9: (i) C-14 isotope is a beta emitter and has a half life of 5730 years. If you take 10gms of C-14, how long it will take to be reduced to 2.5 gms? (ii) You have 1 gm of 92U238 isotope. How long will it take to reduce to 1/16gm? Half life period of U-238 is 4.5 X 109 years. (iii) How much of tritium will be left from 12.8gms of tritium after 7 half life periods? The half life period of tritium s 12.26 years.

CARBON DATING By this method we can know the age of an old object from plant or animal containing carbon. For example, there was excavation (digging the earth) in the Mahenzodaro and Harrappa. During excavation, an old wooden bowl was found. By carbon dating method, it was possible to know the age of this bowl i.e when this wooden bowl was prepared(i.e when the tree from which the bowl was prepared was cut). So indirectly we can know the approximate period before which the Mahenzodaro and Harappan civilization existed. We already know that C-14 isotope is radioactive and is a beta emitter. C-14 decays slowly as its half life period is large i.e 5730 years. C14 ---------> 7N14 + –1e0 6 There are two types carbon dioxide in our atmosphere, i.e C12O2(ordinary) and a very small amount of C14O2(radioactive). C-14 isotope is formed in the atmosphere by the bombardment of neutrons coming from sun(cosmic neutrons) with the ordinary nitrogen(N-14). N14 + 0n1 -------> 1H1 + 6C14 7 A N-14 atom gets converted to a hydrogen atom and C-14 isotope, when a cosmic neutron hits an ordinary atmospheric nitrogen atom. This C-14 atom reacts with oxygen gas to give C 14 O 2 (radioactive carbon dioxide). Plants during photosynthesis take up this C14O2(radioactive)along with the normal C12O2. So C-14 remains present in very small quantities in the carbohydrate that plant prepares during photosynthesis. As long as the plant is living, the C-14 content in the plant remains fixed, since photosynthesis occurs everyday during the day time. Although C-14 is not stable and it is disintegrating by emitting a beta particle, its quantity will not decrease in a living plant because of continued photosynthesis. But as soon as a plant dies, it cannot perform photosynthesis and C-14 content starts to decrease with time as it decays. But we know that the decay of C-14 is very slow as its half life period is 5730 years. So even after thousands of years, the wooden bowl made from a dead plant still would possess radioactivity. Note that the radioactivity of a material is often measured in terms of the number of particles(beta in this case)emitted by one gm of substance per minute. This is done by an instrument called Geiger Muller Counter. In a living plant, whether living now or living thousands of years before(say Mahenzodaro or Harappan period), the radioactivity is fixed(as already mentioned before). It is nearly 15.3 beta particles per minute per gram of the material. This

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means that whether belonging to our time now or during Mahenzodaro's time, one gram of a living plant emits 15.3 beta particle per minute(which means 153 beta particles in 10 minutes). If we measure the beta particle activity in case of the old wooden bowl that we have got during the excavation, we will find that the activity is less than 15.3. Depending on the time passed since the cutting of the tree from which the bowl had been made, the activity of the old object will vary. Suppose, the old wooden bowl emits 9 beta particles per minute per gram now(which we can measure from the Geiger Muller Counter), we can find the age of the bowl (i.e the time when the tree from which the bowl was made died) by comparing the activity of the old bowl with activity of a living plant at that time. We already know that there is no difference between the activity of a living plant now(=a living plant then). By knowing the age of the bowl, we can indirectly know the time when that civilisation existed. In fact, by carbon dating method, the scientists could know the time when old civilizations grew in different parts of the world. For calculating the age of the bowl we shall use the same equation we have used earlier for the radioactivity. The detailed mathematical aspects of these equations will be covered in higher classes. For now, this simple method of calculation will solve the purpose of understanding the concepts.

At = A0 1 2

t T0 .5

Let us take Rt= radioacvitity after time t and R0 =radio activity of a living plant

Rt = R0 1 2

t T0 .5

⇒

1 9 = 15.3 2

t T0 .5

We know that T0.5(half life period) of C-14 is 5730 years. By substituting the value of T0.5 in the above equation, we can solve the equation and find the age of the bowl(t). If you have studied logarithm in mathematics by now, it is very simple to solve this equation by taking log on both the sides. If you have not yet learnt logarithm, then wait until you learn it. After that you take up the calculation part of carbon dating. Now only understand the concepts clearly.

1 log 9 = log 15.3 + log 2

t

5730

⇒ 0.9542 = 1.1847 +

t log 1 2

5730

By solving this equation, we get, t = 4387.92 years. So the age of the wooden bowl recovered from the excavation is 4387.92 years. In other words the civilization existed nearly 4387.92 years before. In this way, the carbon dating is helpful in finding not only the age of an old carbon containing material such as wooden bowl, bone etc. found in an archeological excavation but also knowing the period when an old and perished civilisation grew. Take for example, bones of the dinosaurs species those existed thousands of years ago could be found in many areas of South Africa and American jungles during excavation. The carbon dating method could reveal the period when the dinosaurs species existed. You know that dinosaur does not exist now and many English films have been made on this animal, dinosaur in which the animal is prepared artificially on the basis of the bones found during excavation.

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SAQ 10: A dog femur bone is found during an excavation. The age of the excavated layers is of archaeological interest, so dating by radiocarbon was carried out. The half-life of carbon-14 is 5 730 years. The beta activity of a living carbon-containing matter(living dog) is 15.3 beta particles per minute per gram. The dog femur bone found, showed an activity of 12 beta particle per minute per gram. Calculate the age of the bone in years.

MASS DEFECT AND BINDING ENERGY: When a nucleus is formed, some mass is lost and is converted to its equivalent amount of energy. From the Einstein equation, we can know the relationship between mass loss and energy obtained.

E = (Δ Δm) c2 (where Δm is the mass loss and c is the velocity of light and E is the energy obtained). This energy released when the nucleus is formed is called its binding energy of the nucleus. Let us take the simple example of a 2He4 atom. How much of energy is released when helium nucleus is formed. We know that the nucleus of He contains two protons and two neutrons. The masses of each proton, neutron and electron are known to us in amu scale. Let us find theoretically the atomic mass of He atom. Mass of one proton = 1.0073 amu, Mass of one neutron = 1.0087 amu, Mass of one electron=0.00055amu The atomic mass of He= 2 X 1.0073 + 2X 1.0087 + 2 X 0.00055 = 4.0331 amu(theoretical mass) But actual atomic mass(isotopic mass) of He as obtained by mass spectrometer(please refer the chapter atomic mass for this) is 4.0015 amu. Calculated atomic mass = 4.0331amu, Actual isotopic mass = 4.0015amu Do you find a difference between the calculated (theoretical) mass and actual mass of He atom? Where did this differential mass go? This difference is called Mass Defect. This mass has been lost from the nucleus in the form of energy during its formation. This is called the binding energy of the nucleus. We can calculate this by using the Einstein's equation(E = Δmc2) Δ M = 4.0331 - 4.0015 = 0.0316 amu We know in the chapter, mole concept that 1 amu = 1.67 X 10-24 gm So Ε = Δm c2 = 0.0316 X 1.67 X 10-24 gm X (3 X 1010 cm/s)2 = 4.74 X 10-5 erg This is the binding energy of He nucleus.Binding energy is often expressed in the unit MeV(million electron volt). 1 eV = 1.602 X 10-19 joule =1.602 X 10-12 erg. So 1 MeV = 1.602 X 10-12 X 106 = 1.602 X 10-6 erg So the energy corresponding to the mass defect of 1amu = 1X 1.67 X 10-24gm X (2.988 X 1010 cm/sec)2 = 14.91 X 10-4 erg. Converting the energy from erg to MeV, 14.91 X 10-4 erg = (14.91 X 10-4)/(1.602X10-6) ≈ 931.6MeV So 1 amu ≈ 931.6 MeV We can use this relationship directly to convert mass defect(difference in calculated mass and actual mass) to binding energy in MeV. In case of He atom, we found the mass defect, Δm = 0.0316amu. So the binding energy = 0.0316 X 931 = 29.42 MeV

13

BINDING ENERGY PER NUCLEON: For comparing the stability of different nuclides, binding energy per nucleon is found out. It is obtained when the binding energy of a nuclide is divide by the number of nucleons(protons+neutrons) present in its nucleus. Binding Energy per nucleon = (Binding Energy)/n (where n= number of neutron + no. of protons) In case of He nuclide, we have just discussed, the binding energy per nucleon = 29.42/4 =7.35 MeV SAQ 11: Find the binding energy(B.E) and also binding energy per nucleon(B.E.per nucleon) for (i)7N14 (isotopic mass = 14.0067amu) (ii) 15P31(isotopic mass = 30.9740amu) Mass of one proton = 1.0073 amu, Mass of one neutron = 1.0087 amu, Mass of one electron=0.00055amu Between the two which has a higher binding energy per nucleon.

BINDING ENERGY PER NUCLEON - MASS NUMBER GRAPH :

Binding energy per nucleon versus mass numbers of all natural elements are given in the above diagram. BE/N increases with increase in mass number drastically upto Fe(56) and then decreases graudally with further increase in mass number. Between mass number 40 to 140( Cr, Mn, Fe, Co, Ni, Cu etc) the BE/N values ranges between 8.2 to 8.7 MeV with an average of 8.4 MeV. These constitute the nuclides having intermediate mass numbers which have the highest average BE/N. These nuclides have greater BE/N than lighter nuclides below mass number 40( H, He, C, N, O etc) as well as than heavier nuclides above mass number 140(Pb, Bi, U, Th etc.). For example 92U235 has has BE/N of 7.5 MeV while 2He4 has BE/N of 7.0MeV and 26Fe56 has BE/ N of 8.7 MeV.

14

NUCLEAR ENERGY DURING RADIOACTIVE DECAY: In the previous section, we knew that during the formation of a nucleus there is loss of mass(mass defect) and this mass lost is converted into energy. We shall now see that mass loss also occurs during a radioactive disintegration process. In other words when any radioactive nuclide emits a particle like α, β or a positron, there is always a net loss of mass which is converted to its equivalent amount of energy. The tremendous amount of kinetic energy which the ejected particle possesses is due this energy. Let us take an example. α -emission: When the radioactive nuclide 84Po210 emits an alpha particle, it is converted to 82Pb206 nulclide. We know the actual isotopic masses (experimentally determined) of each of the parent and daughter nuclide and also the mass of an alpha particle. Po210 ----------> He4 + Pb 206 + K.E of α + 0γ0 84 2 82 209.9829amu 4.0026amu 205.9745amu The masses given below each species in the above equation are the actual isotopic masses of the nuclides obtained experimentally. We can find the total mass of RHS and find the difference in mass between LHS and RHS. Mass of RHS = 4.0026+ 205.9745 = 209.9771 amu Mass of LHS = 209.9829, so the loss in mass, Δm = 209.9829 - 209.9771 = 0.0058 amu This amount of mass has been lost when 84Po210 nuclide disintegrates to 82Pb206 by emitting an alpha particle. This mass has been converted to energy as per Einstein's equation. We know that for a mass loss of 1 amu the energy obtained is 931.6MeV, so the energy evolved due to mass loss of 0.0058amu = 931 X 0.0058=5.43MeV. This amount of energy is primarily is used in giving the kinetic energy of the alpha particle and energy of gamm photon which is emitted after an alpha emission. When the ejected particles like alpha, beta or positron hits the surrounding atmosphere, their kinetic energy is converted to heat energy. Like alpha particle, the emission of beta or positron particle involves a mass defect. β -emission: 14

14

C

N + 7

6

0 -1

e+

0

e

+ K.E of β

0

Δm = 0.0017 amu = 0.0017 X 931.6 = 1.58 MeV In this case the decay energy is not sufficient to produce a gamma photon. So in many beta emission processes is accompanied by the emission of another chargeless particle with negligible mass called antineutrino(  ), which is truly anti electron neutrino. So decay energy is used up in giving kinetic energy to the beta particle and forming anti electron neutrino. Positron emssion: 11

C

6

11

0

B + +1 e + 5

0

e

0

In this case too, the decay energy is small like beta emission. Hence it is not sufficient for producing a gamma photon. So many positron emission processes are accompanied by the emission of another chargeless particle having negligible mass called neutrino(  ) which is truly electron

15

neutrino. So decay energy in this case is used up for giving kinetic energy to positron and producing electron neutron. More on neutrinoes will be discussed later.

NUCLEAR FISSION: Neutron induced nuclear fission is commonly understood as nuclear fission. When a heavier radioactive nuclide like uranium-235 isotope(92U235) is bombarded upon by a slow neutron(0n1), the heavier nuclide splits into a pair of lighter nuclides like 56Ba141 and 36Kr92 along with release of 2 to 3 neutrons and a large quantity of energy. This is called nuclear fission and the energy that we get is called the nuclear energy. This is a nuclear reaction but not a nuclear decay process. In spontaneous decay, an alpha or beta or positron is emitted from the nucleus and the daughter nuclide is also a heavy nuclide. It differs from the parent nuclide by maximum of two four units(alpha emission) or no mass unit(beta or positron) and the energy released due to this is small. But in nuclear fission, one subatomic particle, neutron is allowed to hit on a radioactive nuclide like U235 by us. By that it splits the heavy nuclide into two halves forming much lighter daughter nuclides having masses many units smaller than the parent nuclide.The energy released in this process is enormously large. This nulcear energy can be constructively used to get electricity in nuclear power plants and also can be destructively used to prepare nuclear bombs(atom bombs). You know that in India there are many nuclear power plants in places like Kalpakam, Tarapur etc. where electricity is obtained from nuclear power. The heat energy obtained during the fission reaction boils water to produce steam and the steam rotates the turbine kept under a magnetic field to produce electricity. 92

U235 + 0n1 ---> [92U236] --------> 56Ba141 + 36Kr92 + 3 0n1 + Energy

The slow neutron which is bomorbaded upon the U(235) nuclide has speed of 2.2 Km/sec. having energy 0.025 eV. These are also called thermal neutrons because the speed of neutrons at room temperature is nearly 2.2 Km/s. The heavier nuclide absorbs the neutron and is converted to another isotope of uranium(U236). Since this isotope is highly unstable, it immediately splits into two relatively lighter unclides having intermediate mass numbers(explained before) e.g Ba and Kr alongwith 3 neutrons and a large quantity of fission energy. In reality several pairs of daughter nuclides such as Cs/Rb, Xe/Sr, Te/Zr etc are formed alongwith Ba/Kr. But we have given here only one pair (Ba/Kr) for simplicity in understanding.

235 92U

+ 0n1

[ 92U236 ]

90 38Sr

+ 54Xe143 + 3 0n1 +E

92 36Kr

+ 54Ba142 + 2 0n1 +E

52Te

137

144 55Cs

16

+ 40Zr97 + 2 0n1 +E

+ 37Rb90 + 2 0n1 +E

There altogether 300 isotopes(150 pairs of nuclides) from 37 elements which are formed from the fission of U-235 ranging from Zn72 to Dy161. These nuclides are themselves radioactive and are mostly beta emitters. They decay down to 80 stable isotopes belonging to 30 elements. For

example, Kr-92 decays by emitting successive beta particles to form stable istope of Zr-92. 36 Kr

92

-

37 R b

92

t 0.5 = 4.5 s

t 0.5 = 1.84 s

39 Y

92

- t 0.5 = 3.54 h

-

40 Zr

38 Sr

92

- t 0.5 = 2.71 h

92

(stable)

FISSIONABLE AND FISSILE NUCLIDES : The above two terms are often used interchangeably for the same meaning. Actually they convey different meanings. Nuclides which can be fissioned by neutron bombardment are called fissionable nuclides. But out of those which can be fissioned by slow neutrons are called fissile nuclides. Some examples of fissile nuclides are 92U235, 93Np239, 94Pu239, 94Pu241 and 92U233. These are called nulcear fuels. 92U238 is fissionable nuclide but not fissile. It can be fissioned by fast neutrons having speed of 52,000 Km/sec(14.1 MeV), not by slow/ thermal neutrons. Hence all fissile nuclides are fissionable but the vice versa is not true. The other fissionable nuclide which is not fissile is 90Th232. The above two fissionable nuclides namely U-238 and Th-232 are 17

converted to fissile Pu-239 and U-233 respectively by breeding process the details of which will be discussed later. The use of fast neutrons to fission U-238 is not convenient as a chain reaction is not possible for sustained use of the fuel. Hence for all practical purposes these two nuclides are called non-fissile nuclides. But theoretically it is fissionable. Henceforth we shall not use the term fissionable for U-238 and U-232 rather use the term non-fissile for them.

UNCONTROLLED AND CONTROLLED CHAIN REACTION: Natural uranium minerals such as pitchblend, monazite, clevite etc. contain only 0.72% of fissile U-235 nuclide and the rest non-fissile U-298(99.27%), of course a very small quantity of nonfissile U-234(0.005%). When the first slow neutron bombards on the one U-235 nuclide, three neutrons are produced. These neutrons produced are fast neutrons having speed 20,000 Km/sec, but not capable for further fission of both U-235 and U-238. These neutrons are slowed down by using a moderator like light water(H2O) or heavy water(D2O) or graphite. If the % of fissile U-235 i shigh(enriched uranium) then the three neutrons produced in the first fission step bombard on three more U-235 isotopes and each give a pair of Ba and Kr and three additional neutrons + envery. So the bombardment of three neutrons onto three U atoms generates 3 X3 =9 neutrons. These 9 neutrons bombard again on 9 fresh U-235 atoms to generate 9 pairs of daughter nuclides and 9X3=27 neutrons + heat. This chain continues indefinitely. Thus the number of neutrons obtained increases in geometric proportion and the fission reaction propagates in an uncontrolled manner. The time required for the production of a netron in one step and its subsequent bombard on another fissile nuclide as small as 10–6 sec. So within a very short time vast number of fissions take place to generate vast quantity of energy. This is called uncontrolled branching chain reaction. The energy released during the chain reaction is so vast that it is catastrophic and results in an terrible explosion(atom bomb)if the reaction is not controlled. Control of chain reaction is done by using optimising the % fissile nuclide in the nuclear fuel with respect to the moderator used. If natural uranium is used with light water as moderator, then chain reaction cannot build up as most neutrons will be lost without hitting a fissile nuclide. Some hit on abundant U-238 to convert to its heaver isotope which does not fission. Some will escape out of the reactor. So after sometime the fission will stop. U238 + 0n1 -------> 92U239 92 Controlled chain reaction is achieved by properly optimising the percentage of fissile material in the fuel with respect to the moderator used. For light water moderator, enriched uranium containing 2.5 - 3.5% of U-235 is used. India imports enriched uranium ore to feed its light water based nuclear reactors. Natural uranium is used as fuel when heavy water is used as moderator. The optimum mass of fissile nuclide that should be present in the fuel to sustain the chain reaction by allowing only one neutron per fission to undertake further fission and discard the other neutrons is called critical mass of the fissle nuclide. In the side diagram, the fission giving two

18

neutrons per fission has been shown. One neutron is allowed for further fission each time and other neutron is absorbed in the fuel by U-238 or otherwise. Slightly excess to critical mass of fissile nuclide is often taken in the nuclear reactors. Cadmium and boron steel rods are also used to absorb some excess neutrons and thus to control the fission reaction whenever required. These are called controlled rods. When the number of neutron per fission undertaking further fission goes slight above 1, then the control rod is lowered into fuel rods to absorb the excess neutrons. When this number goes slightly below 1, then the control rods are raised away from the fuel rods. To remove the great amount of heat energy produced during fission reaction, coolants are used. Light and heavy water, liquid sodium, or sodium-potassium alloy etc. are used as coolant to absorb the heat produced and remove the heat from the reaction site. Note that the controlled chain reaction continues for years together(usually more than 10 years) in a reactor to get sustained energy in nuclear power plants. Critical, Supercritical and Subcritical Mass : We have already discussed the importance of critical mass of the fissile nuclide for a controlled chain reaction. If the percentage of fissile nuclide is higher than critical mass it is called supercritical mass which goes for a uncontrolled chain reaction as the neutron per fission used for futher fission reactions is above 1. High pecentage of U-235 or pure U-235 is used for the purpose of making nuclear weapons(atom bombs). If the percentage of fissile nuclide is less than the critical value, it is called subcritical mass which goes for stopping the chain reaction after sometime.

Calculation of Fission Energy: (a) Mass defect method: The release of great amount of energy during fission reaction is due to the loss of mass(mass defect) and its conversion into energy. Let us calculate this. The isotopic masses are written below each species in the following equation. U235 + n1 ----------> 37 Rb 93 + Cs 141 + 2 0n1 92 0 55 235.0012amu 92.9035amu 140.894 1 Mass of neutron(0n ) = 1.00885amu Let us find the total mass of LHS which consists of one 92U235 atom and one neutron and total mass of RHS which consists of one 537Rb93, one 55Cs141 and two neutrons. Total mass of LHS : 235.0012 + 1.00885 = 236.01005 amu Total mass of RHS: 92.9035 + 140.894 + 2X 1.00885 = 235.8152 amu We find that the total mass of LHS is more than that of RHS. Thus some mass is lost during the fission reaction. This mass loss has been converted to energy which is called the fission energy. Mass loss = Δm = 2236.01005 - 235.8152 = 0.19485 amu We know that loss of 1 amu mass gives an energy of 931.6 MeV, Energy evolved by one 92U235 atom = 0.19485 X 931.6 = 181.52 MeV Mass evolved by 1 mole i.e 6.023 X 1023 atoms of U = 181.52 X 6.023 X 1023 MeV = 1.093 X 1026 MeV Mass evolved by 1 gm of U atoms = (1/235) X 1.093 X 1026 MeV = 4.651 X 1023 MeV = 4.651 X 1023 X 1.602 X 10–19 X 106 joules. = 7.5 X 1010 J = 7.45 X 107 kJ Thus we find that the fission of 1 gm of U(235) will release such a vast amount of energy. This is incredible!!!

19

(b) Binding Energy Method: The BE/N of the U-235 is 7.6 MeV while the average BE/N of the fragments having intermediate mass numbers(40-140) is 8.4 MeV. This difference of binding energy is released as fission energy during fission reaction. Total binding energy of one U-235 nuclide = 235 X 7.6 = 1786 MeV Taking 37Rb93 and 55Cs141 in the reaction shown above, and using the average BE/N of 8.4 MeV, the total binding, the total binding energy of the fragment pair = 8.4 X ( 93+141)=1965.6 MeV Difference in Binding Energy(Fission energy) per nuclide = 1965.6 – 1786 = 179.6 MeV This value closely matches with the that calculated by using the mass defect method above. This small difference in the results obtained by the above two methods is that in the second method, the average BE/N was used not the exact ones for the pair. Note that the two methods for calculating the fission energy are in principle the same because the mass defect is responsible for the binding energy difference. SAQ 12: What other pairs of daughter nuclides are obtained when U235 is bombarded upon by neutron during fission reaction. SAQ 13: Between the spontaneous alpha emission of U235 and fission reaction by neutron bombardment which process releases greater energy and why?

PRESSURIZED LIGHT WATER REACTOR(PLWR): There are many type of nuclear reactors out of which the common reactors in India are (a) Pressurized Water Reactors(PWR) (b) Boiled Water Reactors (BWR) (c)WWER/ VVER (Water Cooled Water Moderated Energy Reactor) VVER is the Russian abbreviation. In India PWR and VVER are in use. There are two types of PWRs namely (i) Pressurized Lighte Water Reactor(PLWR) and (ii) Pressurized Heavy Water Reactor(PHWR). In India all all the nuclear reactors are of PHWR type and the plant at Kudankulam is of VVER type.To understand the functioning of nuclear reactor easily we shall take up PLWR first.

20

PLWR

PHWR

PLWR : Light water(H2O) is used as coolant-cum-moderator in this reactor which uses enriched uranium ore(2.5-3.5% U-235). It consists of fuel rods installed vertically in the core of a thick cylindrical steel vessel suspended in light water present in the vessel. Each fuel rod is 4 m in length and 1 cm in diameter made of steel or zirconium alloy. This contains the UO2 pellets(enriched), the nuclear fuel. About 200-300 fuel rods are kept together to form one bundle. About 150-200 such bundles are kept in a vertical position inside the cylindrical vessel(1 m diameter adn 1.7 m long) into which the control rods are suspended from upwards. Water acts both as moderator(slowing down the speed of neutron) and also coolant. Since water is kept inside the vessel at 150 atmosphere at 3250C, it is called pressurized water reactor. Water does not vaporize at this perssure and has a much enhanced capacity to carry heat. There are water inlet and exit tubes connected to the vessel. The water carrying the heat produced in the reactor

21

comes out through the exit tube (RHS) and returns back to the to the reactor through the inlet tube after giving away the heat in the heat exchanger. This cyclic rotation of water takes place continuously in the reactor. Water in the exchanger gets converted to steam which runs a turbine which rotates a coil under magnetic field. This electricity is generated. Controlled rods(cadmium rods or boron steel rods) are suspended from above which is made up and down as and when required. When the heat produced becomes more than normal(heading towards uncontrolled chain reaction when neutron per fission goes above 1), the control rods are lowered into the fuel rods so as to reduce the neturons and control the chain reaction. When the heat produced becomes less than the normal(when the number of neutron per fission goes below 1) the control rods are raised up so that more neutrons are allowed to undertake fission and sustain the chain reaction. PHWR: Heavey water(D2O) is used as moderator-cum-coolant in this reactor with unenriched i.e natural uranium as fuel. The basic principles of functioning of PHWR is same as PLWR. Except that the technology is such the reactor can be refuelled without shutting down the reactor. There are a large number of pressure tubes kept in horizontal positions in which fuel is kept as well as heavy water is cycled. We shall not go into the details of PHWR here. In India most of the reactors are PHWR. There are two BWR plants in Tarapur, Rajasthan. Starting of Fission Process : A cource of neutron is brought near the fuel rods in the reactor. The source of neutron consists of a mixture of beryllium powder and an alpha emitter like Ra226 or 98Cf252. Alpha particles bomard on a Beryllium nuclide to form a neutron and a carbon 88 nuclide as shown below. This is called nuclear transmutation about which we shall study later. He4 + 4Be9 → n1 + 6C12 0 Once the fission process starts, the neutron source is removed from the reactor as the reaction becomes self sustaining. 2

FAST BREEDER REACTORS (FBR) In fast breeder reactors, we kill two birds in one stone. We get nuclear energy like PWR/VVER/ BWR, at the same time we get more quantity of fresh fissile nuclide for future use. This is achieved by using highly enriched Uranium(15-30% U-235) and fast neutron to bombard on it. Since breeding of new fissile materials is done here and fast neutrons are used for the purpose, it is called fast breeder reactors. The difference between the conventional nuclear reactors explained before and fast breeder reactors is as follows. (a) FBR uses fast neutrons while conventional reactors use slow neutrons (b) FBR does not use any moderator like LW or HW or graphite while others do. (c) Liquid sodium or Na-K alloy is used as coolant in FBR while in others water is often used as coolant. (d) FBR uses highly enriched(15-30% U-235) uranium while others use either natural or slightly enriched(2.5-3.5%) uranium. FBR consists of two parts in the core of the reactor. (i) Interior part : This part consists of nuclear fuel rods i.e enriched uranium. Fission starts here to generate fresh fast neutrons(at an average of 2.4 per fission) having 20,000 Km/s speed. Out of them, one is used to undertake further fission to sustain the controlled chain reaction and generate sustained energy and the rest 1.4(average) escape out of

22

the interior part to hit on the exterior part.Note that at such high percentage of fissile nuclide, fast neutrons can bring about controlled chain reaction. Slow neutrons are not required for the purpose and hence moderator is not used. Like a conventional reactor, the energy formed in this part produces steam which runs a turbine to generate electricity. (ii) Exterior part : This part surrounds the interior part as a blanket. This part contains natural uranium(0.7% U-235) or depleted uranium( < 0.7% U-235) in the form of rods. The excess 1.4 neutron per fission which escape out of the interior part hit on U238 present in this part to convert finally to 94Pu239 with two successive beta emissions as given below.

92U

238

0n

1

fa st n e u tro n

239

92U

t 0 .5 = 2 3 .5 m in -

-1 e

0

93N p

239

t 0 .5 = 2 .3 5 d a ys -

-1e

0

94P u

239

First U-238 changes to its isotope U-239 by absorbing the fast neutron. U-239 is a beta emitter which changes 93Np239 after emitting a beta particle. The latter is also a beta emitter which changes to 93Np239. The latter two nuclides have small half life periods so as to quickly form the fissile 94Pu239 which has a long half life of 2.44 X 104 years. This fissile 94Pu239 nuclide is stored for a later use in conventional power plants. It has been estimated that for every 100 fissile U235 used up in the interior part for getting nuclear energy 115 fissile Pu-239 are produced in the exterior part for future use. Is it not very interesting !!!! Just like breeding of fissile 94Pu239 is done from U-238, breeding of another fissile nuclide U233 is done from 90Th232 as per the following. 232

90Th

1 0n

233

90Th

-

0 -1e

233 91Pa

-

0 -1e

233 92U

fast neutron In India, commissioning of 500 MW PFBR(Prototype FBR) at Kalpakam will be ready by 2013. 8 more FBRs are on the pipeline at the same place. A 13 MW test FBR is successfully working there since 1985. The 500 MW FBR will make use of blend of Pu-239 and U-235 (their oxide) in uranium spent fuel from conventional plants as fuel. So Pu-239 will be burned in the interior part and depleted uranium will be used as usual in the exterior part to get more Pu-239 in this case. So less Pu-239 is burned and more Pu-239 is breeded. Since the fission of U-233 in nuclear plants is less hazardous,breeding of U-233 from Th-232 is considered as the future of nuclear power in the FBRs.India is the 6th nation to have FBR technology. SOME FREQUENTLY ASKED QUESTIONS(FAQs) ON FISSIONAND THEIR ANSWERS: FAQ 1 : What is spontaneous nuclear fission and how it is different from neutron induced fission. Answer : There are two types of nuclear fission. One is neutron induced which has been discussed in details. The other is spontaneous. Some heavier unstable nuclides having mass numbers much greater than 230 spontaneously fission into two fragments, some neutrons and energy. Cf252

92

→

56

Ba142 + 42Mo106 + 4 0n1 + energy

FAQ 2 : Why it is not convenient to get nuclear energy from U-238 by using fast

23

neutrons. Answer : Fast neutron necessary for fissioning U-238 should have a speed of 52,000 Km/sec. But the neutrons which are obtained from fission, although fast(20,000 Km/s) but not able to undertake further fission. Hence a self sustaining chain reaction is not possible in this case. Hence it is not used in generating nuclear power. FAQ 3 : Why all the fissile nuclides have odd mass numbers ? Answer : Have you marked that all the fissile nuclides carry odd mass numbers like U235, U-233, Pu-239. But nuclides having even mass numbers like U-238 is non-fissile for all practical purposes. The reason is - in nuclide having odd mass number one neutron remains unpaired. When it absorbs another slow neutron, the already existing unpaired electron pairs with the incoming electron and thereby releases some quantity of energy. This is called pairing energy which is 6.8 MeV. Fission of the nucleus needs 6.5 MeV (activation energy) which is less than the pairing energy. Hence fission becomes possible and after fission, we get much more quantity of energy as discussed before. In U-238, there is no unpaired electron. Hence there is no pairing energy in this case. The energy released when it absrobs a neutron is 5.5 MeV which is insufficient to provide the activation energy for fission of U-238 i.e 7.0 MeV. Hence fission is not possible by slow neutrons. But by using very fast neutrons possessing very high K.E, it is possible to overcome the activation energy of U-238 fission. But this is not commercially viable for the reason already discussed. FAQ 4 : When boron steel rods(control rods) are hit by excess neturons what change occurs ? Answer : The following nuclear transmutation occurs when neutron bombards on boron steel. We get an alpha particle alongwith Li-7 isotope. 5

B10 + 0n1

→

3

Li7 + 2He4

FAQ 5 : Where are the heavy water plants present in India ? Answer : There are 7 heavy water plants in India - Nangal, Boroda, Tutikorin, Kota, Thal, Hazira and Manuguru.

NUCLEAR FUSION This is just opposite process of nuclear fission. In nuclear fission that we discussed before, a very heavy nuclide like U235 is broken down into two relatively lighter nuclides(having intermediate mass numbers) and energy is obtained during the process. In case of nuclear fusion, two very light nuclides (eg. 1H1, 1H2 , 1H3) combine or fuse to form relatively heavier nuclides(eg. 2He3, He4) thereby releasing very great amount of energy. This is called fusion reaction and the 2 fusion energy released is much greater than the fission reaction of U235 isotope. 1

H1 + 1H3 → 2He4 + energy

Refer the BE/N – mass number graph given before few pages. The graph rises sharply but falls gradually. Hence binding energy difference between very light nuclides such as protium, deuterium or tritium and the product nuclides like He-3 or He-4 is much greater than that we found for fission reaction. Like fission reaction, the binding energy difference is released, in this case, as fusion energy. We can also explain this on the basis of mass defect. Like the fission reaction, in fusion reaction too, there is a large loss of mass for which the energy released is enormously large. The total mass of the reactants(LHS) is greater than the total mass of the products(RHS)

24

and the difference in mass is lost in the form of fusion energy. Disadvantages in Fusion Reaction : (1) Look to the difference between fission and fusion. In fission, a neutral particle(neutron) was bombarded upon a heavy nuclide like U-235. So there is no difficulty in its absorption by the nucleus. But in fussion, two positively charged nuclei will have to fuse with each other. For that they have to overcome huge repulsive force between them. Therefore this becomes possible by applying very high temperature of the order 107 – 108 0C as well as very high pressure. At high temperature, the kinetic energy of particles would be enormously large and if high pressure is applied simultaneously they would come very close to each other to undergo fusion. So far as temperature is concerned, high temperature is applied initially by external method, but once fusion reaction gets started, the heat produced during the reaction becomes self sustaining and temperature is maintained due to that. No more heat is required from external source. (2) The principal nuclide for fusion is hydrogen. Since it is a combustible gas, there is always a great amount risk involved in using hydrogen. That is why, fusion reactions have not yet been tamed to the extent of using it in a commercial basis. Only it has been a great subject matter of research. But the efforts of the scientists are on to tame down the reaction and once it is achieved, the energy crisis will be all over.

Fusion in Sun and other Stars: The energy produced in sun and stars is due to continuous fusion reactions taking place there. [ 1H1 + 1H1 ---------> 1H2 + +1e0 + 1.44 MeV ] X 2 (step -I) 2 1 3 [1H + 1H ----------> 2H + 5.94 MeV ] X 2 (step-II) 3 3 4 1 H + H --------> H + 2 H + 12.86 MeV (step-III) 2 2 2 1 __________________________________________ 4 1H1 ----------------> 2He4 + 2 +1e0 + 26.72 MeV In the first step two hydrogen (protium) nuclides(protons) fuse to give a deuterium nuclide, a positron and fusion energy of 1.44 MeV. In the second step, the deuterium nuclide formed in the first step fuses with another protium nuclide to form a helium-3 nuclide(2He3) and again fusion energy of 5.94 MeV is released. In the third step, two 2He3 nuclides(formed in the second step)fuse to give one 2He4 nuclide and two protium(1H1) nuclides and again fusion energy of 12.86 MeV. The sum of these three steps can be obtained by cancelling the 1H2 and 2He3 after making necessary multiplications(2 multiplied with each of first and second steps). Thus we find that a great amount of energy(26.72 MeV) is released in all the three fusion reactions taking place in sun. The energy released during the formation of one mole of He-4 from hydrogen in fusion reaction ( 26.72 X 6.023 X 1023 = 160.9 X 1023 MeV) is 9 times greater than the fission energy obtained from 4 gm of U-235. It has been estimated that sun has 1033 gm of hydrogen gas. Only 2% is exhausted in every 109 years.Hence to exhaust all the hydrogen present in the sun, time required will be 1010 years.

Advantages of Fusion Reactions : (1) Nuclear fuel for this reaction is hydrogen, which can be obtained in inexhaustible quantity from sea water in a cheap price compared to Uranium-235 used for fission.

25

(2) The energy obtained from this much greater. (3) It is more safe as very less hazardous radioactive emissions take place from the fusion reactors. Some other Fusion reactions: 1

→ 2He3 + 0n1 + Energy H2 + 1H3 → 2He4 + 0n1 + Energy 1

H2 + 1H2

The fusion of H-2(deuterium) with H-3(tritium) has been possible but deuterim-deuterium fusion has not been possible yet. Because it requires much higher temperature. If this fusion could be possible, then it will be more convenient for getting energy. Deuterium is obtained from ordinary source of hydrogen such as water. But tritium is synthetically prepared with much difficulty as it is not available in nature.

FUSION REACTORS: High temperature in fusion reactors is achieved by (a) microwave furnce (b) laser beam (c) Fas neutron bombardment In Fast Neutron Bombardment(FNB) process, fast neturons are bombarded on hydrogen gas which not only bread the bond and make them atoms but also strip off the electron to convert them to hydrogen ions or protons. This high velocity collision of neutron beam with hydrogen produces also a very hight temperature required for fusion. At such high temperature, all the species exsit as ions. This is called the plasma state of matter.Superconducting magnets are used to compress or apply very high pressure onto the plasma to bring about fusion. Hydrogen bomb: Hydrogen bomb makes use of fusion reaction. Hydrogen bomb is much more devastating than atom bomb which makes use of fission. The fusion reaction involved in the hydrogen bomb is given below. H2 + 1H3 ---------> 2He4 + 0n1 + 17.6 MeV 1 Here one deuterium(1H2) nuclide fuses with one tritium(1H3)nuclide to form one helium(2He4) nuclide and a lot of energy. Hydrogen bomb is also called a thermonuclear bomb. The high temperature needed for the fusion(107 - 108 0C) is obtained by an intial fission reaction. Once fusion starts, it becomes self sustaining. In fact tritium is obtained synthetically by the neutron bombardment on Li-6 nuclide. Therefore for hydrogen bomb, Lithium(6) Deuteride(Li6D) is used to which neutron is bombarded. 6

+

1 0n

3 1H

+

3 1H

+

2 1H

2He

4

3Li

3Li

6

+

2 1H

+ 2He

+

4

1 + 0n

+ Energy + Energy

2 2He4 + 22.4 MeV Energy

Tritium obtained in the first step fuses with deterium to form He-4 and lot of energy. The first hydrogen bomb test by USA in 1952 in Eniwetok island which was based on the above

26

reaction. Now several countries including India has the know-how of a hydrogen bomb.

Artificial Nuclear Transformation or Transmutation and Artificial radioactivity: When highing accelerated subatomic particles like neutron, proton, alpha particles etc.are allowed to bombard onto a nuclide of an element, it is transformed into a nuclide of another element. Thus a new element is produced. This is called nuclear transformation or transmutation. Most of the synthetic elements(transuranic elements, tritum, Tc, Pm, At, Fr, Co-60, C-11, Br-82, Na-24 etc) have been prepared by this technique. The first nuclear transmutation was studied by Chadwick who discovered neutron. When beryllium metal was bombarded upon by alpha particles coming from a radioactive alpha emitter, it was surprisingly found that silvery white beryllium metal was slowly transformed to a black nonmetallic substance, carbon. In this process a beam of neutral particles was emitted. These particles were named as neutrons. Be9 + 2He4 ---------> 0n1 + 6C12 4 Add the mass numbers shown in superscripts in LHS species and RHS species separately you will find them equal. Also add the atomic numbers in subscripts in LHS and RHS which would also be equal. In fact we can know the element formed during the transformation reaction if we know the element onto which bombarded, particle which bombards on the element and the new particle produced in the process. Look to the following process. Let us bombard neutron(0n1) onto 8O16 nuclide. We find that a new particle is produced. This new particle is an alpha(2He4) particle. What is the new element formed in the transformation process? O16 + 0n1 ---------> ? + 2He4 8 Let us add the atomic numbers in the LHS species given in the subscripts. It is 8 + 0 =8. In the RHS, we know the atomic number of alpha particle is 2. So the remaining atomic number in RHS would be 8 – 2=6. Now what is the element having atomic number 6? It is carbon. So carbon is the new element formed. The mass numbers are then added in LHS. It 16 + 1 = 17. In the RHS, an alpha particle has a mass number of 4, so the remaining mass 17 – 4 =13, will be mass number of the carbon formed. So it is 6C13 isotope. O16 + 0n1 ---------> 6C13 + 2He4 8 Thus we find that ordinary gaseous oxygen when bombarded by neutrons is converted to solid carbon(not the usual C12 isotope but the uncommon C13 isotope). This is interesting indeed, just like a magic !! The above reaction shown(oxygen to carbon conversion) is called a 8O16(n,α) reaction. The first particle given inside the bracket(n) is the particle bombarded and the second particle(α) is the particle formed during the transformation process. The target element undergoing transformation is written first. This type of convention is used in all nuclear transformation processes. Transformation by alpha particle: F19 (α,p) reaction : F19 + 2He4 --------> 1H1 + 10Ne22 9 9 33

As75 (α,n) :

Transformation by protons: P31(p,n) reaction : 15

27

→ 35Br78 + 0n1

33

As75 + 2He4

15

P31 + 1H1 ----------> 0n1 + 16S31

13 3

Al27 (p,γ) :

Li7 (p,α) :

→ 14Si28 + 0γ0 Li7 + 1H1 → 2He4 + 2He4 3 13

Al27 + 1H1

Transformation by neutrons: 13 3

Al27 (n,p) :

Li6 (n,T) :

O16 (n,α)

8

Transformation by deuteron: Li6 (d,p) reaction: 3 13

Al27 (d,α) :

→ 12Mg27 + 1H1 Li6 + 0n1 → 1H3 + 2He4 3 O16 + 0n1 → 6C13 + 2He4 ; 8 13

Al27 + 0n1

Deuterium ion (1H2) is called a deuteron. Li6 + 1H2 ---------> 1H1 + 3Li7 3 13

Al27 + 1H2

→ 12Mg25 + 2He4 ;

Transformation by gamma rays: Be9 (γ,n) : Be9 + 0γ0 → 2 2He4 + 0n1 4 Transformation effected by heavier projectiles: For the preparation of many synthetic isotopes 4

most of which are radioactive, heavier nuclides like Li-6, Be-9, B-11, C-12, O-16, F-19, Ne-20 etc. have been used as projectiles to hit still heavier particles. By this method, transuranic elements also have been synthesized.

→ 86Rn205 + 6 0n1 Au197 + 7N14 (projectile) → 35Br74 + 3 0n1 79 U238 + 8O16 (projectile) → 100Fm250 + 4 0n1 92 29

Cu65 + 6C12 (projectile)

SAQ 14: Write the following nuclear equations as indicated. (i) (α,p) reaction of 7N14 (ii) (α,n) reaction of 13Al27 7 (iii) (p,n) reaction of 3Li (iv) (d, n) reaction of 83Bi209 SAQ 15: Fill in the blanks: 50 (i) 24Cr + 2He4 --------> ? + 0n1 (ii) 42Mo96 + ? ---------> 43Tc97 + 0n1 7 1 4 (iii) 3Li + 0n -------> 2 2He + ? (iv) ? + 2He4 --------> 12Mg26 + 1H1 14 1 14 (v) 7N + 0n ------->6C + ?

ARTIFICIAL RADIOACTIVITY: Artificial radioactivity is an usual consequence of nuclear transformation that we studied before. Look to the following (n,p) reaction of 13Al27 giving 12Mg27. Al27 + 0n1 ---------> 1H1 + 12Mg27* (radioactive) 13 You know that the stable isotope of Mg is 12Mg24. So 12Mg27 contains excess neutrons than expected(n/p ratio is more). So this nuclide is not stable and is radioactive. Can you guess what particle it emits? Yes, it is a beta emitter. You know that the nuclides which have greater n/p ratio(which lie above the band of stability) are beta emitters. The beta disintegration reaction of Mg27 is as follows. 12 Mg27 ----------> -1e0 + 13Al27 (Artificial Radioactivity) 12 It decays with a fixed half life period to stable 13Al27 nuclide. This is called artificial radioactivity. Let us take another example of (α,n) reaction of 13Al27 nuclide. Al27 + 2He4 -------> 0n1 + 15P30* (radioactive) 13

28

The 15P30 nuclide produced is not a stable nuclide as P31 isotope is stable. 5P30 has less number of neutrons than required. Since its n/p ratio is less than expected, could you guess what particle it would emit in its radioactivity? Yes, it would emit a positron. The nuclides having lower n/p ratio than expected are positron emitters.So the 15P30 isotope produced is radioactive and decays as follows. P30 -------> +1e0 + 14Si30(stable) (Artificial Radioactivity) 15 SAQ 16: Give the (n,p) reaction of 7N14 and comment on the artificial radioactivity of the product. Give the decay reaction of the product.

RADIOACTIVE SERIES: All the natural radioactive elements and their isotopes belong to four series or families. Each series is named after the first isotope from which the serites originates. For example, in case of uranium series the first istope is U-238 which is an alpha emitter. It forms the daughter element Th-234 after emitting an alpha particle. Th-234 is a beta emitter which forms the daughter nuclide Pa-234. Pa-234 is also a beta emitter which forms another daughter nuclide. This process continues till the last nuclide obtained is a stable one. In case of uranium series it is Pb-206. The half life periods of each radioactive nuclide in the series are different. For exampel U-238 has t0.5 = 4.5 X 109 years while Th-234 has t0.5 = 24.21 days. (1) Uranium Series / (4n + 2) Series :

92U

238

- 

90T h

234

- 

91P a

234

82P b

206

The first nuclide is U-238 and the last stable nuclide is Pb-82. This series contains 14 radioactive nuclides. 8 alpha and 6 beta particles are emitted throughout the series. Since the mass numbers of all the nuclides in this series is expressed as equal to (4n+2), where n is a whole number, it is called (4n+2) series. For example: 238 = 4n + 2, so n = 59(whole number), 206 = 4n + 2, so n = 51.

29

(2)

Thorium Series / 4n Series:

90T h

232

-

88R a

228

-

89A c

228

82P b

208

In this series, the starting nuclide is Th-232 and last stable nuclide is Pb-208. This series contains about 11 radioactive nuclides. At some places branching reactions are found. For example, Bi212 decays both by alpha and beta emission to give two different daughter nuclides which ultimately give the same stable Pb-208 nuclide by emitting beta and alpha particles respectively. Six alpha and 4 beta particles(excluding the branching reaction) are emitted during the series. All

30

the mass number of this series are expressed as 4n where n is a whole number. 232= 4n, so n=58 (3) Actinium Series/ (4n+3) Series:

89Ac

227

82Pb

207

This series begins with Ac-227 nuclide and ends in the stable Pb-207 nuclide. The series has some nuclides above Ac-227. It actually begins from U-235 which is an alpha emiiter and gives Th-231. Th-231 is a beta emitter and produces Pa-231. Pa-231 is an alpha emitter and produces Ac-227 which regarded as the starting nuclide for all practical purposes in this series. Actinium series is also called U-235 series as the true starter is U-235. Since all these nuclides above Ac227 are scarce, it is popularly called actinium series. There are branching reactions at many places in this series. Excluding the braching reactions, 5 alpha and 3 beta particles are emitted in the series. All the mass numbers of this series are express by (4n+3) where n is a whole numberl 227 = 4n +3, So n = 56. 4. Neptunium Series / (4n+1) Series: 93 Np

237

83 Bi

209

In this series, the starting nuclide Neptunium-237 is almost extinct in nature. But all other members are available. The ending stable nuclide is Bi-209.

31

There is branching reaction at one place in this series. Excluding this, the number of alpha and beta particles emitted during the series is 7 and 7 respectively. All the mass numbers are expressed as (4n+1). 237 = 4n + 1, So n = 59. SAQ 16 : Prove that in Uranium series 8 alpha and 6 beta particles are emitted.

Application of radioactivity: The discovery of nucleus from the famous gold foil experiment by Rutherford was possible by the alpha particles emitted from radioactive element. Neutron was discovered by Chadwick by the help of radioactivity. Isotopes and isobars of elements were discovered first with radioactive elements. The artificial elements beyond Uranium(transuranic elements) are the result of radioactivity. The nuclear energy obtained in nuclear power plants is used to get electricity. By carbon dating we could know the age of the old carbon containing material and hence the age of many perished civilisations. Gamma rays are used to cure cancer. Other rays are used to detect many other diseases. The leakage in the underground oil pipe lines, water pipes, gas pipe lines can be detected through a technique called tracer analysis. A radioactive element is introduced in the pipe line and its appearance is detected at suspected places through instruments like Geiger Muller Counter(which detects and counts radioactive emissions). This helps to know the point of leakage. The tracer analysis is also used in hundreds of chemical reactions to study the pathway through which a reaction proceeds. Without dealing with the details, we list out here the areas where radioactivity is used. Diagnosis and Treatment of Diseases: (1)

32

Several radioisotopes are used for diagnosis and treatement of many diseases. I-131 isotope is used for diagnosis and treatment of diseases of thyroid gland. Co-60 is used for cancer, P-32 for leukemia(blood cancer) and brain tumour, Na-24 for heart diseases. Technetium(Tc)-99 is used for taking pictures of heart, lungs, liver, spleen etc. and diagnose diseases. Yttrium(Y)-90 is used for liver cancer while Cu-64 is used to diagnose genetic diseases. (2) Food Industry : Food can be preserved for a long time by using radioactive emission which kill microorganism such as virus, bacteria etc. (3) Preservation of Antique Monuments: By applying the radioactive rays, the antique monuments are being preserved from insects and other microorganism. (4) Industry : Fabric, Paper, rubber, paints industries make use of some radioactive isotopes to determine the quality of the products, thus used in quality control measures. (5) Agriculture : Radioisopes are used with the fertilisers to carry out research on nutrition of plants. The details we shall not discuss here. (6) Age of old organic matter (Carbon dating) : The age of an old civilization could be estimated by carbon dating which has been explained before. (7) Age of Earth: From the ratio of U-238 and Pb-206 present in the uranium minerals, the age of the earth is determined. This called Uranium-lead dating. We shall not discuss more this. (8) Tracer Analysis: already discussed. (9) Sterelization of medical instruments: The equipments and instruments used in hospitals can be sterelized by using gamma radiations obtained from a radioisotope. (10) It helps in maintaining the temperature below earth crust. The radioactive emissions produce a lot of heat when arrested inside earth crust. (11) Nuclear Fission and Fusion: Already discussed before.

Hazards of Radioactivity: Radioactive isotopes and the emitted rays exist in the earth crust as well as atmosphere. The radioactive rays; alpha, beta, gamma are dangerous to human beings. Our food also contain traces of radioactive substances. When these remain below the human tolerance limit, it is not harmful. If it goes above that level can cause dreadful diseases like cancer in blood, skin, breast, liver, colon, bladder, stomach, intestine etc. The radioactive Rn is present much above the limit in the atmosphere of USA and other developed nations which has caused a great concern for them. Human bodies contain radioactive K-40, C-14, Pb-210. Tobacco cotains Pb-210. The flourescent tubes of television contains tritium.A lot of radiactive isotopes and their emissions mix up the atmosphere from the nuclear waste in nuclear power plants, atom bomb test explosions. This has caused a great threat to human civilization and WHO is deeply worried about this

33

nuclear pollution and engaged in combatting it.The persons working in the nuclear plants are more exposed to the harmful effects of the radiations. Precautions should be taken to protect our living environment from the harmful effects of radioactive rays.

Rudimentary Idea on Elementary Particles and Antiparticles: The entire universe is made up of two types of elementary particles. Note that the proton and neutron are also further divisible to still smaller particles. In other words they are made up of elementary particles which are still smaller than them. These two types of elementary particles are (a) QUARKS (b) LEPTONS

QUARKS : There are six types of quarks. (1) UP ( U quark) (2) DOWN (D quark) (3) CHARM (C quark) (4) STRANGE (S quark) (5) TOP (T quark) (6) BOTTOM (B quark) U, C and T quarks carry +2/3 units of charge while D, S adn B quarks carry –1/3 unit of charge.Quarks always remain in groups. Hence they are social beings. LEPTONS: There are 6 types of leptons. (1) ELECTRON (e) (2) MUON (μ) (3) TAU (τ) (4) ELECTRON NEUTRINO (νe) (5) MUON NEUTRINO(νμ ) (6) TAU NEUTRINO (ντ) Each of electron, muon and tau has –1 of charge and the corresponding neutrinos have no charge. So the electron about which you know so much is a lepton. Leptons always remain isolated(not in group), hence they are not social beings. (The beginners are advised not to confuse neutron with neutrino. The are different. While neutron is a baryon made up of quarks, neutrinos are leptons) ANTIPARTICLES: Every particle has its antiparticle. Particle and antiparticle are identical in all properties except one property. For charged particles, particle and antiparticles have opposite charges. Excepting three neutrinos, the antiparticles of all other elementary particles have opposite charges w.r.t their corresponding particles. For example, U quark has +2/3 charge, while anti U quark( U ) has –2/3 charge and so on. The antiparticles are denoted with bar sign above the symbol of their particles. For neutrinos, the antiparticles too have no charge. But they differ in more

34

than one properties, principal among them is difference in masses. We shall not discuss more about this here. Just remember that antiparticle is like the reflection of reflection of a particle which are identical in all respects except charge(valid for charged particles). Note that there are six antiquarks. (1) U (Anti U) (2) D (Anti D) (3) C (Anti C) (4) S (Anti S) (5) T (Anti T) (6) B (Anti B) There are also six antileptons. (1) e (Anti e, commonly called a positron with symbol +1e0) (2)  (Anti μ, this has also +1 charge but has a mass) (3)  (Anti τ, this has also +1 charge but has a mass) (4)  e (Anti electron neutrino, commonly called neutrino. This has no charge but negligible mass like electron) (5)   (Anti muon neutrino. It has no charge but has mass) (6)  (Anti tau neutrio. it has also no charge but has mass) So altogether there 12 particles (6 quarks and 6 leptons) and also 12 antiparticles(6 anti quarks and 6 antileptons). The entire universe is composed of these 12 particle/antiparticle pairs. Antiparticles are unstable and hence do not exist freely. The common antiparticle which is emitted from some radioactive nuclides is positron i.e antielectron. The presence of all other antiparticles have been proved beyond any doubt now. ANNIHILATION: The interesting fact about particle-antiparticle is that when any particle comes in contact with its own antiparticle, both the two are destroyed and their mass is converted to energy. It has been already told that when a positron(antielectron) comes in contact with an electron, both electron and positron destry each other and we get two gamma photons moving in opposite directions. This is called annhilation process. The same is the case with any particle annihilating with its own antiparticle. Just like a particle has its antiparticle, a bigger particle like proton, helium atom etc.i.e a matter has its antimatter. Recently scientists have prepared antiproton, antihydrogen and antihelium atoms. They are extremely unstable having very small life periods. Scientists believe that every matter can have its antimatter, though not available, can be prepared. It is believed that when the universe was formed by the big bang, equal number of matter and antimatter were formed. Immediately after these matter and antimatter underwent annhilation to pure energy. But some matter was left out. In other words there was an imbalance between

35

matter and antimatter after big bang. The universe we see is that left-out matter. So scientists believe that there must be the corresponding antimatter somewhere hidden which we do not find. Often you might have heard the fantasies/humour about this antimatter. Some are excited to believe that there must somewhere an ANTIWORLD where every matter here has an anti version. For example, Anti Sonia must be reading in an anti high school and eating anti ice cream and riding anti cycle and so on. If Sonia living in this world somehow comes in contact with Anti Sonia living in antiworld, can you imagine what will happen ? Both Sonia and antisonia will destroy each other and covert to pure energy !!! This is merely fantasy. HADRONS: Quarks always remain in groups, not isolated. They remain either as a combination of 2 quarks or 3 quarks. These combinations of quarks are called HADRONs. Hadrons are of two types. (1) BARYONS : combination of 3 quarks (2) MESONS : combination of 2 quarks. Examples of Baryons : Proton and Neutron which you know are baryons PROTON : UUD (it is combination of two U quarks and one D quarks. So count the total charge. Is it not +1 ? +2/3 + 2/3 – 1/3 = +1 NEUTRON : UDD (its a combination of one U quark and two D quarks. Is it not netural ? +2/3 – 1/3 – 1/3 = 0 Examples of Mesons: There are many types of mesons out of which the π mesons also called pions are more common. There are three types of pions. (1) π+ (positve pion) : U D (it is combination of one U quark and and anti D quark. (2) π– (negative pion) : D U ( it is a combination of one D quark and an anti U quark). Note that π+ is the antiparticle of π–. (3) π0 (neutral pion) : U U or D D (it is made up of U or D quark with its own antiparticle. Note that there is no annhiliation occuring there as long as a neutral pion is in life. FUNCTION OF PIONS: (1) STABILITY OF NUCLEUS(MESON THEORY): The most important function of pions(pi mesons) is to bind the nucleons(neutrons and protons) to form a stable nuclues. When a neutron emits a π– , it is converted to a proton and when another proton absorbs a π– , it is converted to a neutron. Similarly when proton emits a π+, it is converted to a neutron and when another another neutron absorbs a π+ , it is converted to a proton. Hence the charged pi-mesons rapidly exchange beween proton to neutron and neutron to proton. Neutral pions are exchange between proton to neutron and vice versa. When a proton emits a π0, it remains as a proton and when a neutron absorbs π0 it remains as a nuetron. Theses rapid exhange of pions between the nucleons is responsible for the binding of nucleus and hence its stabililty. In fact, when a U quark changes into a D quark in a proton, it is converted to neutron during beta emission of a radioactive nuclide, (UUD changes to UDD), another particle called W– boson is emitted first, which then decays to a beta particle and an antineutrino. The details

36

about bosons will not be discussed now. D → U + W– W–

→

e0 +  e

–1

(2) Mesons are unstable particles. Pions and other mesons are present in atmosphere formed by the interaction cosmic rays with it. Pions decay into leptons and their neutrinos as follows. There are two modes of the decay of charged pions. Muon emission:

            

Electron emission:

   e  e

   e   e Neutral pion decays by the following two ways

 0  2  0  e  e   GLUONS: Gluons are chargeless and massless force carrier particles which bind the quarks in forming the mesons and baryons. FERMIONS: Particles containing odd number of elementary particles called are spin half particles(do not bother about this term spin half now). Examples are quarks, leptons, baryons. BOSONS: Particles having even number elementary particles are called spin 1 particles(forget about spin 1 now). Mesons are bosons as they contain two quarks. Force carrier particles like photons, gluons, W and Z bosons are also included in this category. FORCE CARRIER PARTICLES: Photons, gluons, W and Z bosons are called force carrier particles. LARGE HADRON COLLIDER(LHC): The mimicry of the big bang is being tried 100 m below ground near Geneva with a 27 Km(circumference) circular gigantic tunnel in the France-Switzerland border surrounding Alps mountain. The protons(handron) are allowed to collide with each other in opposite directions at a velocity close that of light so as to produce elementary particles and their antiparticles, bosons, gluons and other particles. Thus it will recreate conditions that existed during 1st billionth of a second after the big bang. Scientists from more than 100 nations are working for the mega project day and night to prove or disprove the big bang theory of creation of universe. The most important particle they expect to find is HIGG'S BOSON. Higg's boson is the hypthetical particle which was responsible, as they believe, to give mass to massless photons. After more than two LHC Collisions by now, many new findings have been obtained. But scientists are not sure of the presence of Higg's boson yet. If not found, then the big bang 37

theory will go wrong. Some other theory will come up for the creation of the urniverse. The other particles they expect are SUPPER SYMMETRIC PARTICLES. which compose of 96% of universe called DARK MATTER. Astronauts have got the evidence of the existence of dark matter from the unusually high gravitational pull they experience which is not matching with the material universe. LHC experiment consists of the following steps. 1. Linear Accelerator : The hydrogen gas is subjected to electron impact to break it into atoms and strip off one electron from each to covert them to protons. These protons then are allowed to enter the linear accelerator in which they are subjected to an electric field to get velocity nearly one third of light. 2. Four Ring Circular Boosters: The protons are then enter into four circular boosters in which they are subjected to electric and magnetic field. Here they are compressed and get velocity 91.6% of light. Each ring has 157 m circumference. 3. Proton Synchroton: Here again it is subjected to magnetic fields to increase its velocity to 99.9% of light. Note that when a particle travels with velocity nearly equal to that of light, if we give more energy to it, its relativistic mass will increase instead of velocity. Here protons become 25 times heavier than its rest mass. Proton synchroton is circular in shape having circumference of 628 m. 4. Super Proton Synchroton : Again protons are stronger magnetic fields to acquire more energy. Its energy becomes 450 GeV(Giga electron volt). Its mass increase much more. It is also circular in shape having 7 Km circumference. 5 Large Hadron Collider : Ultimately the protons enter into LHC in two counter rotating beams, but not allowed to collide. They are subjected to more energy to acqure a value of 7 TeV(tera electron volt). The proton becomes 7000 times heavier than its rest mass here and rotate 11000 times a second. LHC has 27 Km circumference. After gaining the required energy here they are allowed to collide at 4 positions(Dectector) where the elementary particles, their antiparticles, bosons, gluons, super symmetric particles, Higg's boson etc. are expected to be detected. Time will tell us what is the outcome of the mamoth scientific endeavour. SAQ 17: Can you say, an antiproton is made up of what elementary particles ? How is antihydrogen atom is composed of ?

RESPONSE TO SAQs Radioactivity SAQ 1: (i) Th230 --------> 2He4 + 88Ra226 (ii) Th232--------> 2He4 + 88Ra228 90 90 Both the isotopes of Th i.e 230 and 232 are alpha emitters. Both produce the same elements no doubt(Ra) but they are different isotopes of Ra, one is Ra-226 and the other Ra-228 isotope. (iii) Ra226 --------> 2He4 + 86Rn222 (iv) 86Rn220 --------> 2He4 + 84Po216 88

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SAQ 2: Th234 --------> -1e0 + 91Pa234 (i) 90 We found in the previous SAQ that Th-232 and Th-230 isotopes are alpha emitters while Th-234 we find here is a beta emitter. (ii) K39 -------> -1e0 + 20Ca39 (iii)91Pa234 -------> -1e0 + 92U234 19 60 0 60 (iv) Co -------> -1e + 28Ni (v) 83Bi212 -------> -1e0 + 84Po212 27 238 SAQ 3: 92U is an alpha emitter and its daughter 90Th234 is a beta emitter. We have seen these examples before. There is no relationship between the nature of the emission of parent and daughter nuclides. The parent nuclide can be an alpha emitter while the daughter might be a beta emitter. In some other cases both parent and daughter nuclides can be either alpha or beta emitters. SAQ 4: (i) Mg27 ---------> -1e0 + 13Al27 (ii) Bi210 --------> -1e0 + 84Po210 12 83 213 4 209 (iii) Po ---------> 2He + 82Pb (iv) O15---------> +1e0 + 7N15 84 8 (v) Au200--------> -1e0 + 80Hg200 (vi) Br82--------> -1e0 + 36Kr82 79 35 226 4 222 (vii) Ra --------> 2He + 86Rn (viii) Po210-------> 2He4 + 82Pb206 88 84 38 0 38 (ix) K ---------> +1e + 18Ar (x) I122 --------> +1e0 + 52Te122 19 53 (xi) Fe60 --------> -1e0 + 27Co60 (xii) Np230-----> 2He4 + 91Pa226 26 93 46 0 46 221 (xiii) V ---------> 1e + 22Ti (xiv)87Fr -----> 2He4 + 85At217 23 SAQ 5: Stable nuclides which fall inside the band: Cl-37, K-39, Ne-20, C-13 Unstable and radioactive nuclides: Mg-27: lies above the band(because Mg-24 is stable) P-30: lies below the band(because P-31 is stable), K-40: lies above the band(as K-39 is stable) O-15: lies below the band(as O-16, 17 and 18 are stable), B-10: lies below the band(as B-11 is stable), Br-82: lies above the band(as Br-79 and Br-81 are stable); Fe-60:lies above the band(as Fe-56 is stable); Ag-111: lies above the band(as Ag-108 is stable). SAQ 6: These are beta emitters because their n/p ratio are more than expected(usual). The stable nuclides of these elements are 1H1, 1H2, 35Br79 and 35Br81, 26Fe56, 26Br79,Br81 and 47Ag108. Since the given nuclides have higher mass numbers, they have more number of neutrons. They fall above the band of stability. They get themselves stabilised by emitting a beta particle. H3 --------> -1e0 + 2He3 Br82 -------->-1e0 + 36Kr82, 1 35 Fe60 ---------> -1e0 + 27Co60 Co60 --------> -1e0 + 28Ni60. 27 111 0 111 Ag --------> -1e + 48Cd . 47 SAQ 7: These are positron emitters because, their n/p ratio are less than expected(usual). The stable nuclides of these elements are, P-31, N-14, Ne-20, Na-23, Sc-45, S-32, V-51, Mn-55. But the mass numbers given are less than their respective stable values. So they are positron emitters. By emitting positron, the n/p ratio increases in the daughter nuclides and become more stable. P28 ---------> +1e0 + 14Si28, N12 ----------> +1e0 + 6C12 15 7 18 0 18 Ne --------> +1e + 9F , Na21 --------> +1e0 + 10Ne21 10 11 41 0 40 Sc --------> +1e + 20Ca , S30 ----------> +1e0 + 15P30 21 16 46 0 46 V ---------> +1e + 22Ti , Mn50 --------> +1e0 + 24Cr50 25 23 SAQ 8: Beta emitters: 6C14, 11Na24, 19K40, 12Mg27, 25Mn56, 53I129, 26Co60 Their mass numbers are greater than those of their stable mass numbers given in the SAQ. Positron emitters: 5B10, 11Na22, 19K37, 15P28, 21Sc41, 26

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Their mass numbers are less than those of thier stable mass number values given in the SAQ. Write the products in each case for practice. SAQ 9: (i) For converting 10gm to 5gm(half), one half life period is required. Again for converting 5gms to 2.5gm, another half life period is required. So for the conversion of 10gms of C-14 to 2.5gms of C-14, two half life periods i.e 2X 5730 = 11460 years will be required. (ii) For converting 1gm to 1/2 gm, we need one half life, then for converting 1/2 to 1/4gm we need another half life, then for converting 1/4 gm to 1/8 gm we need one more half life and finally for converting 1/8 to 1/16 gm, we need one more half life period. So a total of four half life periods i.e 4 X 4.5 X 109 years will be required for converting 1 gms of U-238 to 1/16gm. Alternatively:

At = A0 1 2 ⇒

t T0 .5

1/16 = 1 X (1/2)x (where x = t/T0.5= number of half lives) ⇒ (1/2)4 = (1/2)x

⇒

x =4

(four half lives) So the time required = 4 X 4.4 X 109 years. Which method you liked, the second, the formula method or the first, analysis method? (iii) Let us use the formula to get the answer quickly in this case.

At = A0 1 2

t T0 .5

⇒ At = 12.8(1/2)7 = 12.8 X 1/128 = 0.1 gm

(since t/T0.5=number of half life periods) So after 7 half life periods, 12.8gms of tritium will be reduced to only 0.1 gm.(Note that this method is simpler than analysis method, where you have to find the fraction left after 7 half lives and then multiply this fraction with 12.8gms to get the answer). SAQ 10: This would be solved in the same way as shown in the example. The solution is left to you which you can easily do after learning logarithm chapter in mathematics. log(12) = log(15.3) + (t/5730) log (1/2), ⇒ t = 2012 years SAQ 11: (i) N 14: No. of protons =7, No. of neutrons =7 and No. of electrons=7 7 Calculated atomic mass = 7 X 1.0073 + 7X 1.0087 + 7X 0.00055 = 14.11585amu Actual isotopic mass = 14.0067amu(given), So Δm = 14.11585 - 14.0067 = 0.10915amu So binding energy(B.E) of 7N14 = 0.10915 X 931 = 101.62 MeV( Since 1 amu= 931MeV) Binding energy per nucleon(B.E per nucleon) = 101.62/14= 7.26 MeV. (ii) P31: No. of protons = 15, No. of neutrons = 16, No. of electrons= 15 15 Calculated atomic mass= 15 X 1.0073 + 16X1.0087 + 15 X 0.00055 = 31.25695amu Actual isotopic mass = 30.9740amu, So Δm = 31.25695-30.9740 =0.28295amu So binding energy(B.E) = 0.28295 X 931 = 263.42 MeV Binding energy per nucleon(B.E per nucleon) = 263.42/31 = 8.497 MeV. If we compare between the binding energy per nucleon(B.E per nucleon) between the two nuclides, then we find that 15P31 nuclide has greater B.E per nucleon than 7N14 nuclide. SAQ 12: There are 150 pairs (300 isotopes) belonging to 37 elements those are obtained during the fission of U-235. Some pairs are given in the text.

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235 92

U

In these reactions, two neutrons are produced along with the daughter nuclides. SAQ 13: Fission reaction produces much greater energy as the mass loss is more. SAQ 14: (ii) 13Al27 + 2He4 -----> 0n1 + 15P30 (i)7N14 + 2He4 --------> 1H1 + 8O17 (iii) 3Li7 + 1H1 ---------> 0n1 + 4Be7 (iv) 83Bi209 + 1H2 -------> 0n1 + 84Po210 SAQ 15: (ii) 1H2 (deuteron) (iii) -1e0(electron) (iv) (i) 26Fe53 2 Na 11 (v) 1H1 SAQ 16: N14 + 0n1 -------> 1H1 + 6C14**(radioactive) 7 14 Since C has more neutrons than expected(C12 and C13 are stable), it is a beta emitter. C14 --------> -1e0 + 7N14 6 This is the process that takes place in the carbon dating which we have discussed before. SAQ 16 : Let x number of alpha and y number of beta particles are emitted in the uranium series. Let us build up two simultaneous equations from mass numbers and atomic numbers from LHS and RHS. 92

U238 =

82

Pb206 + x 2He4 + y -1e0

238 = 206 + 4x + 0 (1) 92 = 82 + 2x - y (2) On solving, x = 8 and y = 6 You can get these numbers in the same way for other series.

SAQ 17 : Antiproton is made up of two U (anti U) quarks and one D (anti D) quark. So what is the total charge ? Is it not –1 (-2/3-2/3+1/3). In an antihydrogen atom, the neucleus is made up of an antiproton, already discussed and antielectron(positron) would be rotating about the nucleus

IMPORTANT: According to recent convention, the correct representation of a nuclide has both the mass number and atomic number written in the LHS as follows. 235 92

U

So the readers are advised to write according to the above format and not put the mass numbers in the RHS subscript as given in the entire chapter. For saving time, this has been done in this book. However it is not according to recent convention. Please bear with this inconvenience.

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