Solutions To Basic Nuclear Physics (2) ( Nuclear Shell Model) Problems

 1(a)Find the ratio of the helium nucleus to the uranium 238 nucleus.

Solution:

 The ratio of the radii is given by:

 R1/ R2  =  [r _o A1 ^1/3] /  [r _o A2  ^1/3]

 Or:    R1/ R2 = (A1/ A2) ^1/3

 Where: A1 = 4  and A2 = 292

 Then:   R1/ R2 = (4 / 292) ^1/3  

 =  0.239

(b) Estimate, using any technique you can think of, the ratio of the nuclear densities for part (a).

Soln.

The expression for the nuclear density, as a function of the Fermi energy is given by:

 r  =  [2M E_F /  3 ^2/3 p ^4/3 ħ²] ^3/2

 Then the ratio of densities, would be:

 r 1 / r 2  =

 [2M1 E1_{F /}  3 ^2/3 p ^4/3 ħ²] ^3/2/  [2M2 E2_F /  3 ^2/3 p ^4/3 ħ²] ^3/2

 Simplifying:

r 1/r 2   =   [M1 E1_F / M2 E2_F ]  

And:

M2  =  238.05078826 u

M1  =   4.002603254 u 

r 1/r 2   »  0.017 [E1_F / E2_F ]  

Thus, an estimate of the density ratio can be obtained by taking the ratio of the nucleon masses M1 to M2, and multiplying it  by the ratio of the Fermi energies, leaving in symbolic terms of the latter.

2) An element has mass number A = 202 and atomic number Z = 80.

a) Find the diameter of the nucleus and how many times it is greater than that of hydrogen.

Solution:

 R = r _o A ^1/3

 Where:  r _o  =  1.2 x 10 ^-15 m, so D = 2 R

Then: D2  =  2 (1.2 x 10 ^-15 m) (202) ^1/3  =  1.4 x 10 ^-14 m

And the diameter of hydrogen’s nucleus is:

 D1  =  2 (1.2 x 10 ^-15 m) (1) ^1/3  =  1.2 x 10 ^-15 m

So the ratio (which yields how many times the element’s nucleus is larger) is:

D2/ D1 =  11.6

 

b) Find the mass defect D M for this nucleus.

Soln:

 By the regular mass addition formula for nuclei:

 80 (1.008142) + (202 – 80) [1.008982] =  203.747 u

 Then:

DM = 201. 970* – 203.747 = - 1.77 u