## Friday, January 21, 2022

### Solutions to Basic Nuclear Physics Problem (1)

Problem:

Evaluate the terms of the semi-empirical mass formula for the U 238 nucleus, if A = 238 and Z = 92.  From this obtain the mass deficiency DM,  where:

DM   =  ( Z) m p    +  (A- Z) m n   M(Z,A)

Where m  is the proton mass and  m n   the neutron mass

the binding energy  Eb = DM c2 , and also the binding energy per nucleon Eb / A

Solution:

First term:

fo(Z, A) = 1.008142Z + 1.008982 (A – Z)

= 1.008142(92) + 1.008982 (238 – 92) = 240.060

Second term:

f1 (Z, A) = -a1 A = - (0.01692) (238) = -4.02696

Third term:

f2 (Z, A) = +a2 A2/3 =  (0.01912) (238)2/3= 0.734298

Fourth term:

f3 (Z, A) = + a3 (Z2/ A1/3 ) =  (0.00763)[(92)2/(238)1/3 ] = 1.04209

Fifth term:

f4 (Z, A) = + a4 (Z  -  A/2)2/ A = (0.010178) [92 – 119]2/ 238

= 0.311754

Sixth term (e.g. –f(A) since A = 238 is even, and (A – Z) = (238 – 92) is even):

-a5 A– 1/2 = 0.012 (238)– 1/2    =  7.778 x 10-4

Summing these terms up yields: 238.12000 u, but we note the mass of the constituents by the regular mass addition formula for nuclei is:

92 (1.008142) + (238 – 92) [1.008982] = 240. 060436 u

Leading to a mass deficiency (defect) of:

DM = 240. 060436 – 238.12000 = 1.94 u

The binding energy is then:

Eb = DM c2     = (931 MeV/u)(1.94) = 1806.1  MeV

Note here that MeV like eV denotes an energy which is equivalent to 1 million electron volts or:

1 MeV = 10 6 (1.6 x 10-19 J)  =   1.6 x 10-13 J

From this, the binding energy per nucleon can also be obtained:

Eb / A =   1806.1 MeV/ 238 = 7.58 MeV / nucleon