1000 Solved Problems in Modern Physics

438 8 Nuclear Physics – II

8.2.8 Liquid Drop Model ................................

8.33 Deduce that withac= 0 .72 MeV andas=23 MeV the ratiozmin/Ais approx-

imately 0.5 for light nuclei and 0.4 for heavy nuclei.

[Royal Holloway, University of London 1998]

8.34 Determine the most stable isobar with mass numberA=64.

8.35 The masses (amu) of the mirror nuclei^2713 Al and^2714 Si are 26.981539 and

26.986704 respectively. Determine the Coulomb’s coefficient in the semi

emperical mass formula.

8.36 If the binding energies of the mirror nuclei^4121 Sc and^4120 Ca are 343.143 V and

350.420 MeV respectively, estimate the radii of the two nuclei by using the

semi empirical mass formula [e^2 / 4 πε 0 = 1 .44 MeV fm]

8.37 The empirical mass formula (neglecting a term representing the odd – even

effect) isM(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. By finding the min-

imum inM(A,Z) for constantAobtain the expressionZmin= 0. 5 A(1+

0. 25 A^2 /^3 ε/γ)−^1 for the value ofZwhich corresponds to the most stable

nucleus for a set of isobars of mass numberA.

[Royal Holloway, University of London 1998]

8.38 The binding energy of a nucleus with atomic numberZand mass numberA

can be expressed by Weisacker’s semi – empirical formula

B=avA−asA^2 /^3 −

acZ^2

A^1 /^3

−aa

(N−Z)^2

A

−

ap

A^1 /^2

whereav= 15 .56 MeV,as= 17 .23 MeV,

ac= 0 .697 MeV,aa= 23 .285 MeV,ap=− 12 , 0 ,12 MeV

for even-even, odd-even (or even-odd) or odd-odd nucleus respectively.

Estimate the energy needed to remove one neutron from nucleus^4020 Ca.

8.39 (a) Consider the alpha particle decay^23090 Th→^22688 Ra+αand use the follow-

ing expression to calculate the values of the binding energyBfor the two

heavy nuclei involved in this process.

B=avA−asA^2 /^3 −ac

Z(Z−1)

A^1 /^3

−aa

(N−Z)^2

A

−apA−^3 /^4

where values for the constants av,as,ac,aaand apare respectively 15.5,

16.8, 0.72, 23.0 and− 34 .5 MeV. Given that the total binding energy of the

alpha particle is 28.3 MeV, find the energyQreleased in the decay.

(b) This energy appears as the kinetic energy of the products of the decay.

If the original thorium nucleus was at rest, use conservation of momen-

tum and conservation of energy to find the kinetic energy of the daughter

nucleus^226 Ra.