Nuclear Physics-II
PHY-451
Lecture-17
“Nuclear Reactions”
Solved Examples
By: Dr. Sajid Khan
Asst. Professor, Department of Physics
KUST
Topic: Solved Examples on Nuclear Reactions
Outline
Solved Examples on Nuclear Reaction
Identification of the Missing Particle
Calculation of Cross Section
Finding Unknown Mass using Conservation
Laws
Determination of Exoergic and Endoergic
Nature of a Nuclear Reaction
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 2
Topic: Solved Examples on Nuclear Reactions
Identification of the Missing particle
Find the missing nucleus/particle in the following reactions:
𝟓𝟗
𝟏 𝟐𝟕𝑪𝒐 + 𝟏𝟎𝒏 → ? + γ
𝟏𝟒 𝟏 𝟏𝟏
𝟐 𝟕 𝑵 + 𝟏𝑯 → 𝟔𝑪+?
𝟔𝟒 𝟏
𝟑 𝟑𝟎 𝒁𝒏 + 𝟎𝒏 →? + 𝟏𝟎𝒏 + 𝟏𝟎𝒏
𝟐𝟑𝟖 𝟐𝟒𝟑
𝟒 𝟗𝟐𝑼 + 𝟏𝟒𝟕𝑵 → 𝟗𝟕𝑩𝒌 + 𝟓 𝟏𝟎𝒏+ ?
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 3
Topic: Solved Examples on Nuclear Reactions
Identification of the Missing particle
𝟓𝟗
𝟏 𝟐𝟕𝑪𝒐 + 𝟏𝟎𝒏 → ? + γ
Here total number of protons of the reactants is 27 and total number of
nucleons is 60.
Therefore, the resultant nucleus will have 27 protons and 60 nucleons.
So, the missing nucleus is Co and the reaction is
𝟓𝟗 𝟔𝟎
𝟐𝟕𝑪𝒐 + 𝟏𝟎𝒏 → 𝟐𝟕𝑪𝒐 + γ
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 4
Topic: Solved Examples on Nuclear Reactions
Identification of the Missing particle
𝟏𝟒
𝟐 𝟕𝑵 + 𝟏𝟏𝑯 → 𝟏𝟏
𝟔𝑪+?
The missing particle is 𝟒𝟐𝑯𝒆 and the complete reaction is
𝟏𝟒
𝟕𝑵 + 𝟏𝟏𝑯 → 𝟏𝟏
𝟔𝑪 + 𝟒𝟐𝑯𝒆
𝟔𝟒
𝟑 𝟑𝟎𝒁𝒏 + 𝟏𝟎𝒏 →? + 𝟏𝟎𝒏 + 𝟏𝟎𝒏
The missing nucleus is 𝟔𝟑
𝟑𝟎𝒁𝒏 and the complete reaction is
𝟔𝟒 𝟔𝟑
𝟑𝟎𝒁𝒏 + 𝟏𝟎𝒏 → 𝟑𝟎𝒁𝒏 + 𝟏𝟎𝒏 + 𝟏𝟎𝒏
𝟐𝟑𝟖 𝟏𝟒 𝟐𝟒𝟑
𝟒 𝟗𝟐 𝑼 + 𝟕𝑵 → 𝟗𝟕𝑩𝒌 + 𝟓 𝟏𝟎𝒏+ ?
The missing particle is 𝟒𝟐𝑯𝒆 and the complete reaction is
𝟐𝟑𝟖 𝟐𝟒𝟑
𝟗𝟐𝑼 + 𝟏𝟒𝟕𝑵 → 𝟗𝟕𝑩𝒌 + 𝟓 𝟏𝟎𝒏 + 𝟒𝟐𝑯𝒆
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 5
Topic: Solved Examples on Nuclear Reactions
Calculation of Cross Section
A 0.01 mm thick Li target is bombarded with a beam of flux of 1013
particles/cm2-s. As a result 108 neutrons/s are produced. Calculate the cross-
section for this reaction. Given: density of lithium = 500 kg/m3.
Solution:
Thickness of Li = t = 0.01 mm = 10–5 m
Number of 𝟕𝟑𝑳𝒊 nuclei per unit volume =n=ρN/M
=500×6.02×1026/7
=4.302×1028/m3
Number of Li nuclei/area = 4.302 1028 t = 4.302 1028 ×10–5
= 4.302 1023
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 6
Topic: Solved Examples on Nuclear Reactions
Calculation of Cross Section
Number of nuclei undergoing interactions/s = Number of neutrons
produced/s
Number of neutrons produced/s = 108
Number of incident particles striking/unit area of target N0 =
1013/unit area of target.
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒏𝒆𝒖𝒕𝒓𝒐𝒏𝒔 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒅 𝒑𝒆𝒓 𝒔𝒆𝒄𝒐𝒏𝒅
Cross Section = 𝝈 =
𝑵𝟎 ×𝒏
𝟏𝟎𝟖
= 𝟏𝟑
𝟏𝟎 ×𝟒.𝟑𝟎𝟐×𝟏𝟎𝟐𝟑
=2.32× 𝟏𝟎−𝟐𝟗 𝒎𝟐
=0.232 b
(1 b =10-28 m2 )
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 7
Topic: Solved Examples on Nuclear Reactions
Finding Unknown Mass using Conservation Laws
When a proton captures a neutron to produce a deuteron nucleus, a γ-ray
of energy 2.230 MeV is released based on the following equation:
𝟏 𝟏 𝟐
𝟏𝑯 + 𝟎𝒏 → 𝟏𝑯 + γ
Calculate the mass of neutron. Given: Mass of H = 1.008142 amu Mass of 2H =
2.014735 amu
Solution: Total mass of the reactants = Total mass of the
products
Let us take all masses in MeV. Therefore
(1.008142)(931.47) + (m) (931.47) = (2.014735) (931.47) + 2.230
where m is the mass of neutron.
(931.47)(m)=939.8613
Solving for m = 1.008987 amu
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 8
Topic: Solved Examples on Nuclear Reactions
Exoergic and Endoergic Nuclear Recations
The Q value of a nuclear reaction can be positive, negative, or zero.
If Q > 0 (minitial> mfinal or Tfinal> Tinitial) the reaction is said to be exoergic or
exothermic.
In this case nuclear mass or binding energy is released as kinetic energy of
the final products.
When Q < 0 (minitial< mfinal or Tfinal< Tinitial) the reaction is endoergic or
endothermic.
In this case initial kinetic energy is converted into nuclear mass or binding
energy.
The changes in mass and energy must of course be related by the familiar
expression from special relativity, ∆E = mc2.
Any change in the kinetic energy of the system of reacting particles must be
balanced by an equal change in its rest energy.
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 9
Topic: Solved Examples on Nuclear Reactions
Exoergic and Endoergic Nuclear Reactions
Find out whether the following reaction is exoergic or endoergic.
6Li + 1n 4He + 3H
Given: M(6Li) = 6.0151234 amu M(1n) = 1.0086654 amu M(4He) = 4.0026034
amu M(3H) = 3.0160294 amu
Q = [M(6Li) + M(1n) – M(4He) – M(3H)]
= [6.0151234 + 1.0086654 – 4.0026034 – 3.0160294]
= 0.005156 amu = 0.005156 931.47 = 4.8 MeV
Therefore, the Q-value for this reaction is 4.8 MeV and the reaction is exoergic.
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 10
Topic: Solved Examples on Nuclear Reactions
END OF LECTURE
Questions can be asked on KCMS or through
official email address
Course: Nuclear Physics-II, PHY-451, Instructor: Dr. Sajid Khan, Asst. Prof. Department of Physics, KUST, Email: sajidkhan@kust.edu.pk 11
