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 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
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.
Minus
the liquid drop and Coulombic repulsion corrections noted above, the region of
highest stability shown is simply the region for which Eb / 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]
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 EF / 3 2/3 p 4/3 ħ2]
3/2
Then the ratio of
densities, i.e. r1 / r2 would
be:
r1/r2 =
[2M1 E1F / 3 2/3 p 4/3 ħ2]
3/2/ [2M2 E2F /
3 2/3 p 4/3 ħ2]
3/2
Simplifying:
r1/r2 = [M1
E1F / M2 E2F ]
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: Eb = DM
c2 = (931 MeV/u)(1.77) =
1647.8 MeV
From
this, the binding energy per nucleon is:
Eb
/ A = 1647.8 MeV/
202 = 8.15 MeV / nucleon
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