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 _p  is the proton mass and  m _n   the neutron mass

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

Solution:

First term:

 f_o(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 A^2/3 =  (0.01912) (238)^2/3= 0.734298

Fourth term:

f₃ (Z, A) = + a₃ (Z²/ A^1/3 ) =  (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: 

E_b = 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:

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