**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:

f

_{1}(Z, A) = -a_{1}A = - (0.01692) (238) = -4.02696
Third
term:

f

_{2}(Z, A) = +a_{2}A^{2/3 }= (0.01912) (238)^{2/3}= 0.734298
Fourth
term:

f

_{3}(Z, A) = + a_{3 }(Z^{2}/ A^{1/3 }) = (0.00763)[(92)^{2}/(238)^{1/3 }] =
1.04209

Fifth
term:

f

_{4}(Z, A) = + a_{4 }(Z^{ }- A/2)^{2}/ A = (0.010178) [92 – 119]^{2}/ 238
=
0.311754

Sixth
term (e.g.

**since A = 238 is even, and (A – Z) = (238 – 92) is even):***–f(A)*
-a

_{5}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 c

^{2 }= (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

**can also be obtained:***binding energy per nucleon*
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.

**.**

*Ans**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

» 50 _{b}/ A = max is effectively constant over:__<__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:

-ħ

-ħ

^{2}/ 2m ( d**y**

^{2}**/dx**

^{2}) = E**y**

**Thence:**

d

d

**y**

^{2}**/dx**

^{2}+ 2m/ ħ^{2}(E**y**

**) = 0**

And the quantized energy is found to be:

E

_{n}= (h^{2}/ 8m L^{2}) n^{2}
This
can be modified to yield a simplified

**in one dimension by using instead:***Fermi energy*
E

_{n}= (h^{2 }p^{2}/ 8m L^{2}) (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

**if we assume its shells filled up to N =4 and use an estimate for L as***nucleus**R = r*_{o}A^{1/3}

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

*Solution:*
We have: E

_{n}= (h^{2 }p**/ 8m L**^{-2}^{2}) (N/2)^{2}
= 4[(6.62 x 10

^{-34}J-s)^{2}(3.14)^{ 2}/ (8m L^{2})
Where L is obtained from:

*R = r*_{o}A^{1/3}
= (1.2 x 10

^{-15}m) (16)**= (1.2 x 10**^{ 1/3}^{-15}m) 2.5 = 3.0 x 10^{-15}m
Then:

E

4[(6.62 x 10

_{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}ħ^{2}]^{ 3/2}

Then the ratio of
densities, i.e.

*r**1 /**r***would be:***2*

*r*

*1/*

*r*

*2 =*

*[2M1 E1*_{F / }3^{2/3}

*p*

^{4/3}ħ^{2}]^{ 3/2}/ [2M2 E2_{F }

_{/ }

*3*^{2/3}

*p*

^{4/3}ħ^{2}]^{ 3/2}
Simplifying:

*r*

*1/*

*r*

*2 = [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:

**= 1.2 x 10***r*_{o}^{-15}m, so D = 2*R*
Then:
D2 =
2 (1.2 x 10

^{-15}m) (**= 1.4 x 10***202)*^{1/3}^{-14}m
And
the diameter of hydrogen’s nucleus is:

D1 = 2 (1.2
x 10

^{-15}m) (**= 1.2 x 10***1)*^{1/3}^{-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 c

^{2 }= (931 MeV/u)(1.77) = 1647.8 MeV

From
this, the

**is:***binding energy per nucleon*
E

_{b}/ A = 1647.8 MeV/ 202 = 8.15 MeV / nucleon
## No comments:

Post a Comment