1000 Solved Problems in Modern Physics

460 8 Nuclear Physics – II

Hence the rest mass energy of the nucleus

M(A,Z)c^2 =Zmpc^2 +(A−Z)mnc^2 −B

= 92 × 938. 3 + 143 × 939. 5 − 1 , 786

= 218 ,929 MeV

Therefore the mass is

218 , 929

931. 5

= 235 .028 amu

The plot is based on the semi-empirical mass formula obtained in the Liquid

Drop Model. In this formula the lowering of binding energy at low mass num-

bers due to surface tension effects as well as at highZ(and hence largeA) due

to coulomb energy, are predicted by this model. The jumps in the curves at low

mass numbers (A=2–20) are attributed to the shell effects explained by the

shell Model. The asymmetry term occurring in the mass formula is explained

by the Fermi Gas Model. From the plot the binding energy per nucleon for^87 Br

is found to be 8.7 MeV and that for^145 La it is 8.2 MeV. The energy released

in the fission is

Q=[B(Br)+B(La)]−B(U)

= 8. 7 × 87 + 8. 2 × 145 − 1786

=160 MeV

8.41 The liquid drop model gives the value ofZfor the most stable isobar of mass
numberAby

Z 0 =

A

2 + 0. 015 A^2 /^3

ForA= 127 , Z 0 = 53 .38, the nearest beingZ=53. Hence^12753 I is stable.

But^12754 Xe is unstable againstβ+decay ore−capture.

8.42^2712 Mg−^2713 Al=− 0 .000841 (12–13)+ 0. 0007668 × 27 −^1 /^3 (12^2 –13^2 )

+ 0. 09966

[(

12 −

27

2

) 2

−

(

13 −

27

2

) 2 ]

=+ 0 .225331 amu

As the right hand side is positive^2712 Mg is heavier than^2713 Al, and therefore it is

unstable againstβ−decay.

8.3.9 Optical Model ....................................

8.43 Introduce the complex potentialV=−(U+iW) in the Schrodinger’s equation

∇^2 ψ+

2 m

^2

(E+U+iW)ψ=0(1)