Selected Solutions: Intro. to Nuclear Physics (Pt. 1)

1.  Evaluate the terms of the semi-empirical mass formula for the U 238 nucleus, if A = 238 and Z = 92. Use this information to find the total mass in atomic mass units (u) and compare it to the standard mass expression. Thence or otherwise, obtain the mass defect and the binding energy as well as the binding energy per nucleon.

 

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 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

 

2.The sketch graph shown  (See Blog Post – Questions, for reference) plots the mass number A vs. the binding energy per nucleon (BE/nucleon) on the vertical axis.

 Using any models or other explanations, account for the region of greatest stability indicated. 

 

Ans.

 Appeal may be made to the “liquid drop” nuclear model. The diagram shown actually shows a constancy of E_b / A  beyond the A  »22 value and beyond. I.e. if a correction is made at smaller A of this range for surface effects – analogous to the surface tension in a water droplet – and a correction is also made at larger A for Coulomb repulsion (of protons) then the main portion of the binding energy per nucleon curve (E_b / A    vs. A) is found to be nearly constant.

Minus the liquid drop and Coulombic repulsion corrections noted above, the region of highest stability shown is simply the region for which E_b / A = max is effectively constant over:

»  50 < A <  75

 

Without applying either of the corrections.

3. A quantum square well potential is defined according to (see 'square well' displayed in question (3) in Intro. To Nuclear Physics post): 

It is found that the Schrodinger equation to solve becomes:

-ħ²/ 2m ( dy²/dx² ) = Ey

Thence:

dy²/dx² + 2m/ ħ² (Ey) = 0

And the quantized energy is found to be:

E _n = (h²/ 8m L²) n²

 

This can be modified to yield a simplified Fermi energy in one dimension by using instead:

E _n = (h²  p²/ 8m L²) (N/2)²

 

Where all the Fermi shell energy levels are presumed occupied up to N/2.  Use this to obtain a simplified estimate of the energy associated with the oxygen nucleus if we assume its shells filled up to N =4 and use an estimate for L as  R = r _o A ^1/3

(Take m = 1.7 x 10 ^-27 kg)

 

Solution:

We have:   E _n = (h²  p^-2/ 8m L²) (N/2)²

 

=  4[(6.62 x 10^-34 J-s)² (3.14) ²  /  (8m L²)

Where L is obtained from: R = r _o A ^1/3

 

=  (1.2 x 10^-15 m) (16) ^1/3   =   (1.2 x 10^-15 m) 2.5 = 3.0  x 10^-15 m

Then:

E _n  = 

4[(6.62 x 10^-34 J-s)² (3.14) ²  / [ (8 (1.7 x 10 ^-27 kg)( 3.0  x 10^-15 m)²]

 

E _n  =     1.4 x 10 ^-10 J 

 

5.(a) Find the ratio of the helium nucleus radius to that of 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).

 

Ans.

 

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, i.e. r1 / r2  would be:

r1/r2   =

 

 [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:

r1/r2   =   [M1 E1_{F /} M2 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.

 

6. 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.

Solution:

 

By the regular mass addition formula for nuclei:

 

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

 

DM = 201. 970* – 203.747 = - 1.77 u

 

(* From table of atomic masses)

 

c)  Solution:

The binding energy is then: E_b = DM c²     = (931 MeV/u)(1.77) = 1647.8  MeV

 

From this, the binding energy per nucleon is:

E_b / A =   1647.8   MeV/ 202 = 8.15 MeV / nucleon