Brief Intro to Nuclear Chemistry 1. What is nuclear decay? A decay in which an unstable nucleus transforms into a stable nucleus through a series of “decays”. A decays is the loss of various components from the nucleus including alpha and beta particles, gamma rays, positrons, etc. 2. What determines the stability of a nucleus? The ratio of protons to neutrons in the nucleus determines the nucleus stability. 3. What is one purpose that the neutrons serve in the nucleus? They separate the positively charged protons from one another – thereby minimizing the repulsions felt within the nucleus. 4. What are some methods of decay? a. α particle decay. An alpha particle is equivalent to a Helium atom.

b. β particle decay. A beta particle is equivalent to an electron.

c. Positron emission Is basically, a positively charged electron (anti-matter).

d. Gamma ray emission. High energy that is released in some decays.

e. Electron capture. This occurs in heavy elements. These elements have a nucleus full of

positively charged protons, because of this, the core electrons are highly attracted to (and therefore very close to the nucleus). Because of the very tight proximity electrons may “fall” into the nucleus. In very simplified terms, the electron that was “captured” and a proton in the nucleus, convert into a neutron. 5. Balance the following reactions:

a.

b.

c.

d. 6. What is a decay series? this is when an unstable nucleus has to go through a number of decays in order to reach a stable configuration. For example:

7. The isotope 247Bk decays by a series of α particle and β particle emissions, eventually ending up as 207Pb. In the complete decay series, how many α and β particles are produced? The mass indicates the number of α particles needed: 247-207 = 40/4 = 10 α particles required. The charge indicates the number of β particles needed: 97 = 10(2) + x(-1) + 82
x = 5 β particles required

8. What is a half-life? This is the amount of time that it takes a sample to decrease to ½ of its initial value. We will only be looking at nuclear decays in terms of being first order. The equation for a first order decay is:

9. Bismuth-210 is radioactive and decays by β particle production and has a half-life of 5 days. a. How much of a 1.00-g sample of 210Bi is left after 2 weeks? t1/2 = 5 days = 0.693 / k
k = 0.1386 days-1 ln N = - (0.1386 days-1) (14) + ln (1)
N = 0.14 g 210 Bi b. How long does it take for 75% of a sample of 210Bi to decay? ln (0.25) = - (0.1386 days-1)(t) + ln(1.00)
t = 10 days 10. What force holds the nucleus together? The Strong Force. 11. Where is the energy for this binding force derived? From the nucleons (neutrons and protons). Remember that based on the quantum mechanical model, when you are down to the subatomic level, the difference between matter and energy is very grey. So when a nucleus combines some of the proton and neutron characteristics are used as energy to bind the nucleus together and the remaining portion is translated as mass. This leads to a different mass for a nucleus, once the particles have combined, than anticipated based on the mass of the individual protons and neutrons. 12. What do we look at to determine the binding energy? We look at the “mass defect” – the difference between the expected mass based on the total number of proton and neutrons in the nucleus and the actual measured mass.

13. What equation can be used to relate mass energy? E = mc2 14. The sun radiates 3.9 x 1023J of energy into space every second. What is the rate at which mass is lost from the sun?

15. The most stable nucleus in terms of binding energy per nucleon is 56Fe. If the atomic mass of 56Fe is 55.9349 amu. Calculate the binding energy per nucleon for 56 Fe. (1 amu = 1.66054 x 10-27 kg) mn = 1.00866 amu

mp = 1.00728 amu

me = 5.48580 x 10-4 amu

The binding energy depends on the mass defect. So the first thing we have to do is determine ∆m. ∆m = “expected nuclear mass” – “actual nuclear mass” “expected nuclear mass” nuclear mass of 56Fe = 26 (mass of a proton) + 30 (mass of a neutron) mass = 26 ( 1.00728 amu) + 30 (1.00866 amu) = 56.4491 amu “actual nuclear mass” An important aspect of this question is that we are given the actual atomic mass. Though we frequently treat electrons as though they are mass deficient, they actually do contribute a small portion to he overall mass of an atom. However, electrons do not contribute to the binding energy that holds the nucleus together. This means when we deal with these kinds of problems we are only concerned with the mass of the nucleus. Thus means that when we are given the atomic mass of a substance we need to subtract the contribution that the electrons make. actual nuclear mass of 56Fe = 26 (mass of an electron) mass = 55.9349 amu – 26 (5.48580 x 10-4 amu) = 55.9209 amu

The question asked for units of energy per nucleon. The nucleon value is equal to the total number of protons and neutrons in a nucleus. In the case of 56Fe there are 56 nucleons.

16. What is fission? Fission is the breaking of a heavier nucleus into smaller nuclei. 17. What is fusion? Fusion is the combining of smaller nuclei to create a heavier nucleus.