Nuclear Decays and Reactions Xinhua Bai Last updated on February 12, 2015

Introduction 1. Nuclear decay

• Stable nuclei only occur in a very narrow band in the Z − N plane. All other nuclei are unstable and decay spontaneously in various ways. • Isobars with a large surplus of neutrons gain energy by converting a neutron into a proton via β− -decays . • In the case of a surplus of protons, the inverse reaction may occur: i.e., the conversion of a proton into a neutron via β+ -decays Four decay chains on next pages: Thorium, Neptunium, Uranium (Uranium-238), and Actinium (Uranium-235). 1

2

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2. General properties of decay process In a collection of identical unstable nuclei the number of decays per unit time is

dN(t) = −λN(t)dt

(1)

where λ is the decay constant or decay rate. From Eq. (1),

N(t) = N(0)e−λt ln2 τ1/2 = λ

(2) (3)

The mean lifetime is:

R ∞  dN(t)  − dt tdt 0 τ = R ∞  dN(t)  − dt dt 0

(4) (5)

Use Eq. (1) and Eq. (2) to get

1 τ1/2 = λ ln2 where τ1/2 is the halflife of the decay process. τ=

(6)

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Multimodal decay of 164 Ho

dN(t) = −[λ1 N(t) + λ2 N(t)]dt (7) N(t) = N(0)e−(λ1 +λ2 )t (8) λ = λ1 + λ2 (9) 1 1 τ= = (10) λ λ1 + λ2 where λ is the total decay constant; τ is the total lifetime. The branching fractions are:

f1 =

λ1 , λ

f2 =

λ2 λ

(11)

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Sequential decay: 218

Rn → 214 Po → 210 Pb . dN1 (t) = −λ1 N1 (t)dt dN2 (t) = λ1 N1 (t)dt − λ2 N2 (t)dt N1 (t) = N1 (0)e−λ1 t  λ1  −λ1 t −λ2 t N2 (t) = N1 (0) e −e λ2 − λ1

(12) (13) (14) (15)

This can be extended to systems involving arbitrary numbers of sequential decays. See a technical note at http://www.phy.sdsmt.edu/˜bai/2015Spring/PHYS-433/AA_DecayReview2012_

XB_JL.pdf

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Alpha Decay: A XZ →A−4 YZ−2 +4 He2

1. Energy & Momentum Assuming the parent nucleus is at rest, the conservation of energy is nuclear mass M , kinetic energy T :

MX c2 = MY c2 + T Y + Mα c2 + T α total kinetic energy:

(16)

T Y + T α = (MX − MY − Mα )c2 using atomic mass M(A, Z):

(17)

T Y + T α = [M(A, Z) − M(A − 4, Z − 2) + M(4, 2)]c2 ≡ Q

(18)

Q is called the disintegration energy, or simply Q-value of the decay. 7

For non-relativistic case, kinetic energies:

1 1 T Y = MY v2Y , T α = Mα v2α 2 2 momentum conservation: Mα MY vY = Mα vα , vY = vα MY

(19)

(20)

We are more interested in the energy, momentum, etc. of the α particle. Let’s eliminate vY .

1 1 T Y + T α = MY v2Y + Mα v2α 2 2 !2 1 Mα 1 T Y + T α = MY vα + Mα v2α 2 MY 2 ! 1 Mα T Y + T α = Mα v2α +1 2 MY ! Mα Mα + MY TY + Tα = Tα + 1 = Tα MY MY

(21)

In terms of the Q-value (which we know from the atomic masses): Q ≡ T Y + T α = [M(A, Z) −

M(A − 4, Z − 2) + M(4, 2)]c2

Tα =

MY Q Mα + MY

TY = Q − Tα = Q −

(22)

MY Mα Q= Q Mα + MY Mα + MY

(23)

We have the following conclusions:

• When a heavy parent nucleus decays, most of the kinetic energy is carried by the α particle. • Tα =

MY Mα +MY Q

is unique, a result of a two-body decay from a parent particle at rest.

But things may become more complicated ...

Kinetic energy spectrum of alpha particles emitted from 228 Th90 .

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The explanation: Internal energy levels in the daughter nucleus.

Are these real? - Introducing nuclear experiment techniques: (1) The measurement of alpha, gamma particles. (2) Measurement of lifetime. (3) Coincidence technique.

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2. How can this happen?

α

E  ~  5  MeV  

10

The Coulomb potential energy:

ZZ 0 e2 V(r) = ke r ke = 8.99 × 109 Nm2C −2 , R = 1.2A1/3 f m,

[theCoulomb0 sconstant]

[Eq.(2.16)]

for example 228 Th90 :

90 × 2 × (1.6 × 10−19 C)2 ZZ 0 e = ke V(r) = ke r 1.2 × 2281/3 f m 8.99 × 109 Nm2 90 × 2 × 2.56 × 10−38 C 2 = · C2 1.2 × 6.1 × 10−15 m = 5.66 × 102+9+15−38 J = 5.66 × 10−12 ×

1 MeV 1.60217657 × 10−13

V(r) ≈ 35 MeV  5 MeV. The α decay is a quantum effect, call the Barrier Penetration.

2. Barrier Penetration

( V(x) = The Schrodinger equation

~2 d2 ψ(x) − 2m dx2

      

d2 ψ(x) dx2 d2 ψ(x) dx2

0 V0

f or x < 0 f or x > 0

(24)

+ V(x)ψ(x) = Eψ(x) becomes: + k2 ψ(x) = 0 − κ2 ψ(x) = 0

f or x < 0 f or x > 0

(25) 11

where

!1/2 2m k= 2E ~ " #1/2 2m κ = 2 (V0 − E) ~ When 0 < E < V0 , the solutions (eigenfunctions) are

By requiring ψ(x) and finite!), one get

ψ(x) = Aeikx + Be−ikx ,

f or x < 0

ψ(x) = Ceiκx + De−iκx ,

f or x > 0

/ to be finite & continuous for all values of x (because

dψ(x) dx

( ψ(x) = where α = 2tan

−1

  1/2  V0 − E −1 .

2Aeiα/2 cos(kx − α2 ), 2Aeiα/2 cos α2 e−κx ,

f or x < 0 f or x < 0

/

d2 ψ(x) dx2

is

(26)

We use a semi-classical treatment to find out the order of magnitude of the α decay probability. Using the effective potential (bottom), the transmission coefficient T is

T=

4k1 k (k1 +k)2



 κ2 −k k 2 

1 1 + 1 + κ(k1 +k) sinh2 2κa #1/2 " 2Mα (E + U0 ) k1 = ~2 " #1/2 2Mα k= E ~2 " #1/2 2Mα κ= (V0 − E) ~2

(27)

(28) (29) (30)

• Pay attention to the quantities on the two plots to the right: V0 , 2a, . . . • Typical values, V0 ≈ 14 MeV , 2a ≈ 33 f m for Z ≈ 90. 12

Mα : the rest mass of α particle; E : the kinetic of the emitted α particle (observed in the decay, far away from the barrier) ; A numerical example: Mα c2 = 4000 MeV , E = 4.05 MeV , V0 = 14 MeV , U0 = 40 MeV , we have

i1/2 1 h 2 κ= 2Mα c (V0 − E) ~c ≈ 1.4 f m−1 k ≈ 0.9 f m−1 k1 ≈ 3.0 f m−1 2κa ≈ 33 f m × 1.4 f m−1 = 46.2 ! 2κa 2 e 1 = e92  1. sinh2 (2κa) ≈ 2 4 sinh2 (2κa) is the dominant factor in determining the T .

(31)

(32)

Taking the limits k12  κ2 , k12  k2 ,

T=

T T T T

4k1 k (k1 +k)2

  κ2 −k k 2  1 1 + 1 + κ(k1 +k) sinh2 2κa

4κ2 k 1 ≈ 2 κ + k2 k1 sinh2 2κa !1/2 E 4(V0 − E) ≈ · 4 · e−92 V0 E + U0 !1/2 4(10) 4 · 4 · e−92 ≈ 14 44 ≈ 3.5 × e−92 ≈ 4 × 10−40

(33)

How could this tiny transition probability be observed? We have to take into account how often the α particle bounce on the barrier, N -times per second. The decay probability is P = N × T - A semi-classical treatment.

We know the (maximum possible) kinetic energy of the α particle inside the nucleus (following the Fermi gas model) is,

Eα ≈ U0 + E = 44 MeV The corresponding velocity is, r r 2T α 2 × 44 MeV vα = ≈c ≈ 0.15c. Mα 4000 MeV The rate if bouncing again the barrier is, vα 0.15 × 3 × 1010 cm/s ≈ N= R 7.4 × 10−13 cm N ≈ 6 × 1021 s−1

(34)

(35)

(36)

Numbers are for the decay 232 Th →228 Ra +4 He : R = 1.2 × 10−13 A1/3 cm ≈ 7.4 × 10−13 cm. So, the probability of α-emission P (or called the decay constant, λ) is

P≈

vα T ≈ 2.4 × 10−18 s−1 = λ R

(37)

The mean lifetime for the decay process τ is

τ=

1 1 10 ≈ s = 1.3 × 10 year. P 2.4 × 10−18

(38)

The measured α-decay lifetime of 232 Th is 1.4 × 1010 year. Two useful approximations: (1) For V0  E , the decay constant can be written as 4a

1/2

P(α − emission) ∝ E 1/2 e− ~ [2Mα (V0 −E)]

(39)

(2) For V0  E and (V0 − E)1/2 varying slowly with E , we have

logP(α − emission) ∝ logE 1/2 + constant.

(40)

It tells us the relationship between the decay constant and the energy of the decaying particle. It is known as the Geiger-Nuttual Rule. See the plot on the next page (of mean lifetime τ ∝ P1 !).

The  Geiger-­‐Nu,all  rela/onship  between  the  alpha-­‐ decay  halflife  and  the  decay  energy  for  some  even-­‐Z   nuclei.   Note: Requirements from the conservation of the angular momentum and the parity. See an

example decay:

237

Np (5/2+ ) →233 Pa (3/2− ) + α

• Angular momentum conservation requires the angular momentum of the α particle l takes ~ values: JD − J~P ≤ l ≤ J~D + J~P . So, l = 1, 2, 3, 4. • Parity conservation requires the α particle to have odd parity (P = −1). Therefore, l = 1, 3.