1000 Solved Problems in Modern Physics

454 8 Nuclear Physics – II

Therefore (a^2 −b^2 )= 14 .79 fm^2 (1)

Now,A=

4

3

πab^2 ρ (2)

The nuclear charge density,ρ= 0 .17 fm−^3 (3)

Using the value ofρandA=181 in (2), we get

ab^2 = 254 .3fm^3 (4)

Solving (2) and (4) we finda= 7 .1fm,b= 6 .0fm

8.3.5 Nuclear Stability...................................

8.17 GivenB/A= 9. 402 − 7. 7 × 10 −^3 A
For the parent nucleus
B(A,Z)= 9 .402A− 7. 7 × 10 −^3 A^2

For the product nucleus

B(A− 4 ,Z−2)= 9 .402(A−4)− 7. 7 × 10 −^3 (A−4)^2

B(∝)= 28. 3

Condition that alpha decay is just energetically possible is

B(A,Z)=B(A− 4 ,Z−2)+B(∝)

Or 9. 402 A− 7. 7 × 10 −^3 A^2 = 9 .402(A−4)− 7. 7 × 10 −^3 (A−4)^2 + 28. 3

Simplifying and solving forA, we find thatA=153. Thus, alpha decay is

energetically possible forA>153.

8.18 Sn−Sp= 15. 7 − 12. 2 = 3. 5 =

3

5

×

e^2

4 πε 0 R

[Z^2 −(Z−1)^2 ]

Substitutee^2 / 4 πεo= 1 .44 MeV fm andZ=8tofindR= 3 .7fm.

8.19 An atom of massM 1 will decay into the product of massM 2 andαparticle of
mass mαifM 1 >M 2 +mα. Now the mass excessΔ=M−A.
For^229 Th,M 1 = 229 + 0. 031652 = 229 .031652 amu. Forαdecay, the prod-
uct atom would be^225 Ra, andM 2 = 225 + 0. 023528 = 225 .023528.
Forα- particle,mα= 4 + 0. 002603 = 4. 002603
M 2 +mα= 229. 026131
Since,M 1 >M 2 +mα,^229 Th will decay viaαemission.
Forβ−decay,M 2 = 229 + 0. 032022 = 229. 032022
AsM 1 <M 2 , decay viaβ−emission is not possible. By the same argument
the decay of^229 Th to^229 Ac is not possible viaβ+emission asM 1 <M 2 + 2 me.