This document discusses nuclear binding energy and nuclear models. It begins by explaining how the mass of an atom is calculated based on its nuclear mass and electron mass. It then discusses how the actual nuclear mass is less than the sum of its nucleon masses due to binding energy. Greater binding energy leads to more stable nuclei. The binding energy per nucleon varies with atomic mass number A and peaks around iron. Light nuclei can undergo fusion to form heavier nuclei, releasing energy. Heavy nuclei can undergo fission to form lighter nuclei, also releasing energy. Separation energies reflect the stability of removing particles from nuclei. Nuclear spins and dipole moments are also discussed.

4. (Binding Energy & Nuclear Models)

5. (2-1) Nuclear Binding Energy

6. Since an atom contains Z positively
charged particles (protons) and N=A-
Z

7. neutral particles (neutrons), the total
charge of a nucleus is +Ze, where e
represents

8. the charge of one electron. Thus, the
mass of a neutral atom, Matom, can be
expressed

9. in terms of the mass of its nucleus, M
and its electrons me.

10. nuc

11. where m is the proton mass, m the
mass of an electron and m the mass of
a neutron.

12. p

13. e

14. n

15. For example the mass of the rubidium
nucleus,

16. 50 neutrons, can be calculated as:

17. 87

18. Rb, which contains 37 protons and

19. The atomic mass, indicated on most
tables of the elements, is the sum of
the

20. nuclear mass and the total mass of the
electrons present in a neutral atom. In
the case

21. of

22. 87

23. Rb, 37 electrons are present to balance
the charge of the 37 protons. The
atomic

24. 87

25. mass of Rb is then:

26. From the periodic table, the
measured mass of a

27. measured 87

28. MA ( Rb) = 86.909187 amu.
These two masses are not equal and the
difference

29. is given by:

30. 87

31. Rb atom is found to be

33. Expanding the terms in this
equation, shows that the difference
in mass

34. corresponds to a difference in the mass
of the nucleus

35. which reduces to

36. Thus, when using atomic mass
values given by the periodic table,
the mass

37. difference between the measured and
calculated is given by

38. Notice also that

39. Where m is a mass of the hydrogen
atom.

40. H

41. From this and other examples it can
be concluded that the actual mass of
an

42. atomic nucleus is always smaller than
the sum of the rest masses of all its
nucleons

43. (protons and neutrons). This is because
some of the mass of the nucleons is
converted

44. into the energy that is needed to
form that nucleus and hold it
together. This

45. converted mass, Δm, is called the
“mass defect” and the corresponding
energy is

46. called the “binding energy” and is
related to the stability of the nucleus;
the greater

47. binding energy leads to the more
stable the nucleus. This energy
also represents the

48. minimum energy required to
separate a nucleus into protons and
neutrons. The mass

49. defect and binding energy can be
directly related, as shown below:

50. B(A,Z) m 931.5MeV / amu

51. or

52. B(A,Z) 931.5(Zm Zm
Nm Mmeasured

53. )

54. p

55. e

56. n

57. atom

58. Since the total binding energy of
the nucleus depends on the
number of nucleons, a

59. more useful measure of the
cohesiveness is the average binding
energy Bave.

60. B(A,Z)

61. B (A,Z)

62. ave

63. (MeV / nucleon)

64. A

66. Figure (2-1): Variation of binding
energy per nucleon with the atomic
mass number

67. The binding energy per nucleon
varies with the atomic mass
number A, as

68. shown in figure (2-1). For
example, the binding energy per
nucleon in a rubidium

69. nucleus is 8.7MeV, while in
helium it is 7.3MeV. The curve
indicates three

70. characteristic regions:

71. Region of stability: A flat
region between (A) equal to
approximately 35 and 70

72. where the binding energy per
nucleon is nearly constant. This
region exhibits a

73. peak near A = 60. These nuclei are
near iron and are called the iron
peak nuclei

74. representing the most stable
elements.

75. Region of fission reactions:
From the curve it can be seen that
the heaviest nuclei

76. are less stable than the nuclei
near A = 60, which suggests that
energy can be

77. released if heavy nuclei split apart
into smaller nuclei having masses
nearer the

78. iron peak. This process is called
fission (the basic nuclear reaction
used in atomic

79. ____________________________
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81. bombs as uncontrolled reactions
and in nuclear power and
research reactors as

82. controlled chain reactions). Each
fission event generates nuclei
commonly

83. referred to as fission fragments
with mass numbers ranging from
80 to 160.

84. Region of fusion
reactions: The curve of binding
energy suggests a second

85. way in which energy could be
released in nuclear reactions.
The lightest

86. elements (like hydrogen and
helium) have nuclei that are
less stable than

87. heavier elements up to the iron
peak. If two light nuclei can
form a heavier

88. nucleus a significant energy
could be released. This process is
called fusion,

89. and represents the basic
nuclear reaction in hydrogen
(thermonuclear)

90. weapons.

91. (2-2) Separation energy

92. Are the analogous of the
ionization energies in atomic
physics, reflecting the

93. energies of the valence nucleons.
The separation energy of any
particle is defined as

94. the amount of energy needed to
remove a particle from the nucleus.
For a given N,Z;

95. S , S is larger for nuclei with
even N or Z than with odd one,
this due to the pair

96. n

97. p

98. effect of nuclear force which
increase the binding energy and sep
uparation energy.

99. There are two equations to
determine the separation energy
using the mass or the

100. binding energy. For neutron as
follow:

101. S =931.5[M(A-1,Z)+m -M(A,Z)]

102. n

103. n

104. S =B(A,Z)-B(A-1,Z)

105. n

106. For proton:

107. S =931.5[M(A-1,Z-1)+m -
M(A,Z)]

108. p

109. H

110. S =B(A,Z)-B(A-1,Z-1)

111. p

112. For alpha particle:

113. S =931.5[M(A-4,Z-2)+m -
M(A,Z)]

114. α

115. α

116. S =B(A,Z)-B(A-4,Z-2)-B(α≡ 2

117. 4

118. He )

119. α

120. The fact, that each pair of these
equations represents two equivalent
equations.

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122. Nuclear Physics; Ch.2 (Binding
Energy & Nuclear Models); Dr.
Ali A. Ridha

127. Example: calculate the separation
energy of neutron for

128. methods, where M(209
)=209.05398u, M(208

129. 209

130. Pb by using the two

131. )=208.04754u.

132. 82

133. 82

134. Sol.:

135. 1- S =931.5[M(A-1,Z)+m -
M(A,Z)]= 931.5[M(208 )+m -
M(

136. 209

137. 82

138. )]

139. n

140. n

141. 82

142. n

143. =931.5[208.04754+1.008665-
209.05398]=2.07259MeV

144. 2- B( A,Z ) 931.5( Zm
Zm Nm M
measured

145. )

146. p

147. e

148. n

149. atom

150. 209

151. 82

152. B(

153. )=931.5(82x1.007276+82x0.00054
9+127x1.008665-209.05398]

154. =1572.48844MeV

155. 208

156. 82

157. B(

158. )=931.5(82x1.007276+82x0.00054
9+126x1.008665-208.04754]

159. =1570.41585MeV

160. S =B(A,Z)-B(A-1,Z)=1572.48844-
1570.41585=2.07259MeV

161. n

162. Figure (2-2): Neutron separation
energy of lead isotopes as a
function of neutron

163. number.

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165. Nuclear Physics; Ch.2 (Binding
Energy & Nuclear Models); Dr.
Ali A. Ridha

166. 16

168. (2-3) Nuclear Spins and Dipole
Moments

169. 1

170. Both the proton and the
neutron have spin angular
momentum of .

171. 2

172. Furthermore, just as electrons in an
atom can have orbital angular
momentum, so also

173. can nucleons inside a nucleus.
We know from quantum
mechanics that orbital

174. angular momentum can take on
only integral values. The total
angular momentum of

175. the constituents-namely, the
vector sum of the orbital and
intrinsic spin angular

176. momenta-defines the spin of the
nucleus.

177. Thus, the nuclei with even atomic
number have integral nuclear spin
whereas

178. nuclei with odd atomic number
have half-integral nuclear spin.
However, the nuclei

179. with an even number of protons
and an even number of neutrons
(even-even nuclei)

180. have zero nuclear spin. These
facts lend credence to the
hypothesis that spins of

181. nucleons inside a nucleus are very
strongly paired so as to cancel their
overall effect.

182. To explain the fine structure of
the spectral lines, suppose that
each of the

183. electron, proton and neutron has
spin momentum result from
rotation on its axis and

184. therefore they has a magnetic
moment due to this rotation, the
interaction of magnetic

185. moment of the electron with the
magnetic moment of the nucleus
leads to increase or

186. decrease the tension between
them and then will increase or
decrease energy of

187. electron, i.e. split the energy levels
of the electron and thus will be
divided every line

188. of spectral lines into several
lines. Every charged particle has
a magnetic dipole

189. moment associated with its spin,
given by:

190. e

191. gs

193. s

194. s

195. 2mc

196. where e, m and s are the charge,
mass and the intrinsic spin of the
charged

197. particle. The constant g is known as
the Lande factor (Gyromagnetic
ratio), which for

198. a point particle, such as the
electron, is expected to have the
value g = 2.

199. When g ≠ 2 the particle is said
to possess an anomalous magnetic
moment,

200. which is usually ascribed to the
particle having a substructure
like proton and

201. neutron:

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203. Nuclear Physics; Ch.2 (Binding
Energy & Nuclear Models); Dr.
Ali A. Ridha

204. 17

206. Relate to angular momentum

207. e

208. g

211. 2mc

212. g 1 for proton

214. g 0 for neutron

216. 1

217. For the electron (with s ) ,
the dipole moment μ ≈μ , where μ
is the Bohr

218. e