22.02 Problem Set 5 Solution
Krane 5.1 Solution
Let’s take extreme limit of the shell model: assume only the single unpaired nucleon determines
the properties of the nucleus.
(a)
7
Li: A=7, Z=3, N=4
There is an unpaired proton at 1p
3/2
. Spin: 3/2, Parity: (1)
1
=1.
Thus I
π
=3/2

(b)
11
B: A=11, Z=5, N=6
There is an unpaired proton at 1p
3/2
. Spin: 3/2, Parity: (1)
1
=1.
Thus I
π
=3/2

(c)
15
C: A=15, Z=6, N=9
There is an unpaired neutron at 1d
5/2
. Spin: 5/2, Parity: (1)
2
=1.
Thus I
π
=5/2
+
(d)
17
F: A=17, Z=9, N=8
There is an unpaired proton at 1d
5/2
. Spin: 5/2, Parity: (1)
2
=1.
Thus I
π
=5/2
+
(e)
31
P: A=31, Z=15, N=16
There is an unpaired proton at 2s
1/2
. Spin: 1/2, Parity: (1)
0
=1.
Thus I
π
=1/2
+
(f)
141
Pr: A=141, Z=59, N=82
There is an unpaired proton at 2d
5/2
. Spin: 5/2, Parity: (1)
2
=1.
Thus I
π
=5/2
+
Krane 5.2 Solution
13
C: A=13, Z=6, N=7
In the
ground
state, all nucleons stay at the lowest possible energy level as follows:
1d
5/2
1d
5/2
1p
1/2
1p
1/2
1p
3/2
1s
1/2
1p
3/2
1s
1/2
6 protons
7 neutrons
The unpaired single neutron stays at 1p
1/2
, which gives I
π
=1/2

First excited state: I
π
=1/2
+
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22.02 Problem Set 5 Solution
1d
5/2
1d
5/2
1p
1/2
1p
1/2
1p
3/2
1s
1/2
1p
3/2
1s
1/2
6 protons
7 neutrons
The unpaired single neutron stays at 1s
1/2
, which gives I
π
=1/2
+
Second excited state: I
π
=3/2
+
1d
5/2
1d
5/2
1p
1/2
1p
1/2
1p
3/2
1s
1/2
1p
3/2
1s
1/2
6 protons
7 neutrons
The unpaired single neutron stays at 1p
3/2
, which gives I
π
=3/2

Third excited state: I
π
=5/2
+
1d
5/2
1d
5/2
1p
1/2
1p
1/2
1p
3/2
1s
1/2
1p
3/2
1s
1/2
6 protons
7 neutrons
The unpaired single neutron stays at 1d
5/2
, which gives I
π
=5/2
+
Krane 5.3 Solution
203
Tl: A=203, Z=81, N=122
We only consider the single unpaired nucleon in this problem. Check figure 5.6, we find the 81
st
proton should stay at 1h
11/2
, which means I
π
=11/2

for
203
Tl. If the pairing force increases strongly
with bigger l, it is more favorable in terms of lowering energy for a proton to stay at 1h
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 Spring '09
 Mucciolo
 mechanics

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