1000 Solved Problems in Modern Physics

8.3 Solutions 459

8.36 The difference in the binding energies of the two mirror nuclei is assumed to
be the difference in the electrostatic energy, the mass number being the same.

B(Ca)−B(Se)=

3

5

[Z(Z+1)−Z(Z−1)]e^2

4 πε 0 R

e^2 / 4 πε 0 = 1 .44 MeV fm,andZ= 20

The left hand side is 350. 420 − 343. 143 = 7 .227 MeV, Ris calculated as

4.75 fm.

8.37 The empirical mass formula is

M(A,Z)=Z(mp+me)+(A−Z)mn−αA+βA^2 /^3 +γ(A− 2 Z)^2 /A+εZ^2 A−^1 /^3

whereα,β,γandεare constants. HoldingAas constant, differentiateM(A,Z)

with respect toZand set∂∂MZ =0.

∂M

∂Z

=mp+me−mn− 4 γ(A− 2 Z)/A+ 2 εZA−^1 /^3 = 0

The termsmp+me−mn∼=mH, the mass of hydrogen atom which is neglected.

Rearranging the remaining terms, we obtain

Zmin=

A

2 +(ε/ 2 γ)A^2 /^3

which is identical with the given expression.

8.38 B

( 40

20 Ca

)

= 15. 56 × 40 − 17. 23 × 402 /^3 −

0. 697 × 202

401 /^3

− 0 +

12

401 /^2

(1)

B

( 39

20 Ca

)

= 15. 56 × 39 − 17. 23 × 392 /^3 −

0. 697 × 202

391 /^3

−

23. 285

39

−0(2)

Subtracting (2) from (1) gives us the binding energy of neutronB(n) =

15 .38 MeV.

This is the energy needed to separate one neutron from the nucleus.

8.39 (a)B(Th)= 15. 5 × 230 − 16. 8 × 2302 /^3 − 0. 72 ×

90 × 89

2301 /^3

− 23. 0 ×

502

230

+

34. 5

2303 /^4

= 1743 .70 MeV

B(Ra)= 15. 5 × 226 − 16. 8 × 2262 /^3 − 0. 72 ×

88 × 87

2261 /^3

− 23. 0 ×

482

226

+

34. 5

2263 /^4

= 1740 .88 MeV

B(α)= 28. 3

Q=B(Ra)+B(∝)−B(Th)

= 25 .48 MeV

(b)E(Ra)= 4 Q+× 2264 = 0 .44 MeV

8.40 ForA=235 the diagram (Fig. 8.7) indicates the binding energy per nucleon,
B
A

= 7 .6 MeV. Therefore, the total binding energy of^235 U nucleusB =

7. 6 A= 7. 6 × 235 =1,786 MeV.