An Introduction to Nuclear Physics

Chapter I: nuclear massess and binding energies

by Dr. Jiangming Yao,

Nuclear theory and nuclear astrophysics group,

School of Physics and Astronomy, Sun Yat-sen University

Basic formulas

  • Nuclear (atomic) mass:

$$ M(Z,A) = ZM_H + (A-Z)M_n,\quad M_H\approx M_p + m_e $$

  • Nuclear (atomic) mass excess:

$$ \Delta M(Z,A) = M(Z,A) - A m_u, $$ where $m_u$ is defined based on the mass of $^{12}$C,

$$ m_u = M(Z=6, A=12)/12 = 931.493856 {\rm MeV}. $$

  • Nuclear binding energy:

$$ B(Z, A) = Z M_H + (A-Z)M_n - M(Z, A) $$

Numerical calculation for $^{16}$O

In [13]:
mass_excess_O16=-4.737 # MeV

BE_12C=7.680144*12 # MeV

M_H = 938.272+0.511 # MeV
M_p = 938.272 # MeV
M_n = 939.565 # MeV
In [19]:
# atomic mass for 12C
Mass_12C = 6*M_H+6*M_n - BE_12C

# 1 u
Unit=Mass_12C/12 

print("Atomic Mass for 12C: {} [MeV]".format(Mass_12C))
print("1 u : {} [MeV]".format(Unit))

print("Mass excess for 12C : {} [MeV]".format(Mass_12C-12*Unit))

Atomic Mass for 12C: 11177.926272 [MeV]
1 u : 931.493856 [MeV]
Mass excess for 12C : 0.0 [MeV]
In [20]:
# atomic mass for 16O

Mass_16O = Unit * 16 + mass_excess_O16

print("Atomic Mass for 16O: {} [MeV]".format(Mass_16O))

Atomic Mass for 16O: 14899.164696000002 [MeV]
In [18]:
# binding energy for 16O

BE_16O = 8*M_H+8*M_n - Mass_16O

print("Binding Energy for 16O: {} [MeV]".format(BE_16O))
print("E/A for 16O: {} [MeV]".format(BE_16O/16))

Binding Energy for 16O: 127.61930399999801 [MeV]
E/A for 16O: 7.976206499999876 [MeV]

Binding energies

In [40]:
import pandas as pd

import matplotlib.pyplot as plt
In [35]:
df=pd.read_csv('./dataset/AME2003.DAT',sep='\s+',skiprows=None)
df.columns=['A','Z','Mass_excess']
In [36]:
df.head()
Out[36]:
A Z Mass_excess
0 9 6 28.91050
1 10 6 15.69870
2 11 6 10.65030
3 12 6 0.00000
4 13 6 3.12501
In [37]:
df['N']=df['A']-df['Z']
In [38]:
df['BE']=df['Z']*M_H+df['N']*M_n - (Unit * df['A']  + df['Mass_excess'])
In [39]:
df.head(30)
Out[39]:
A Z Mass_excess N BE
0 9 6 28.91050 3 39.037796
1 10 6 15.69870 4 60.320740
2 11 6 10.65030 5 73.440284
3 12 6 0.00000 6 92.161728
4 13 6 3.12501 7 97.107862
5 14 6 3.01989 8 105.284126
6 15 6 9.87310 9 106.502060
7 16 6 13.69400 10 110.752304
8 17 6 21.03900 11 111.478448
9 18 6 24.93000 12 115.658592
... ... ... ... ... ...
20 18 7 13.11400 11 126.692592
21 19 7 15.86200 12 132.015736
22 20 7 21.77000 13 134.178880
23 21 7 25.25000 14 138.770024
24 22 7 32.04000 15 140.051168
25 12 8 32.04800 4 58.549728
26 13 8 23.11200 5 75.556872
27 14 8 8.00736 6 98.732656
28 15 8 2.85560 7 111.955560
29 16 8 -4.73700 8 127.619304

30 rows × 5 columns

In [43]:
plt.scatter(df['A'],df['BE']/df['A'],c='k')
plt.xlabel('Mass number A',fontsize=12)
plt.ylabel('E/A [MeV]',fontsize=12)
plt.legend(['AME2003'])
Out[43]:
<matplotlib.collections.PathCollection at 0x143999fd0>
Out[43]:
Text(0.5,0,'Mass number A')
Out[43]:
Text(0,0.5,'E/A [MeV]')
Out[43]:
<matplotlib.legend.Legend at 0x1439a3588>

References: