Chapter 13 Nuclei Solutions
Question - 21 : - From the relation R = R0A1/3,where R0 is a constant and A isthe mass number of a nucleus, show that the nuclear matter density is nearlyconstant (i.e. independent of A).
Answer - 21 : -
We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0 = Constant.
A = Mass number of the nucleus
Nuclear matterdensity, 
Let m be the average mass ofthe nucleus.
Hence, mass of the nucleus = mA
Hence, the nuclear matter density isindependent of A. It is nearly constant.
(positron) emission from a nucleus, there is another competingprocess known as electron capture (electron from an inner orbit, say, theK−shell, is captured by the nucleus and a neutrino is emitted).Show that if emission is energetically allowed,electron capture is necessarily allowed but not vice−versa.
Answer - 22 : -
Let the amount of energy released during theelectron capture process be Q1. The nuclear reaction canbe written as:
Let the amount of energy released during thepositron capture process be Q2. The nuclear reaction canbe written as:
= Nuclear mass of 
= Nuclear mass of 
= Atomic mass of
= Atomic mass of
me = Mass of anelectron
c = Speed of light
Q-value of the electron capture reactionis given as:
Q-value of the positron capturereaction is given as:
It can be inferred that if Q2 >0, then Q1 > 0; Also, if Q1>0, it does not necessarily mean that Q2 > 0.
In other words, thismeans that ifemission is energetically allowed, then theelectron capture process is necessarily allowed, but not vice-versa. This isbecause the Q-value must be positive for an energetically-allowednuclear reaction.
Question - 23 : - In a periodic table the average atomic mass of magnesium is given as24.312 u. The average value is based on their relative natural abundance onearth. The three isotopes and their masses are 
(23.98504u), 
(24.98584u) and 
(25.98259u). The natural abundance of is 78.99% by mass. Calculate the abundancesof other two isotopes.
Answer - 23 : -
Average atomic mass of magnesium, m =24.312 u
Mass of magnesiumisotope, m1 = 23.98504 uMass of magnesiumisotope, m2 = 24.98584 uMass of magnesiumisotope, m3 = 25.98259 uAbundance of, η1= 78.99%Abundance of, η2 = x%Hence, abundance of, η3 = 100− x − 78.99% = (21.01 − x)%We have the relation for the average atomicmass as:
Hence, the abundance of is 9.3% and that of is 11.71%.
Question - 24 : - The neutron separation energy is defined as the energy required toremove a neutron from the nucleus. Obtain the neutron separation energies ofthe nuclei 
and 
from the following data:
= 39.962591 u
) = 40.962278 u
= 25.986895 u
) = 26.981541 u
Answer - 24 : - For 
For
A neutron 
is removed from a nucleus. The corresponding nuclear reaction can be written as:
It is given that:
Mass = 39.962591 uMass =40.962278 uMass 
= 1.008665 uThe mass defect of this reaction is given as:
Δm = 
∴Δm =0.008978 × 931.5 MeV/c2
Hence, the energy required for neutronremoval is calculated as:
For , the neutron removal reaction can be writtenas:
It is given that:
Mass = 26.981541 uThe mass defect of this reaction is given as:
Hence, the energy required for neutronremoval is calculated as:
Question - 25 : - Under certain circumstances, a nucleus candecay by emitting a particle more massive than an α-particle. Consider thefollowing decay processes:
Calculate the Q-values for thesedecays and determine that both are energetically allowed.
Answer - 25 : - Take a 
emission nuclear reaction:
We know that:
Mass of 
m1 = 223.01850 uMass of 
m2 = 208.98107 uMass of, m3 = 14.00324 uHence, the Q-value of thereaction is given as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 208.98107 − 14.00324) c2
= (0.03419 c2) u
But 1 u = 931.5 MeV/c2
∴Q = 0.03419 ×931.5
= 31.848 MeV
Hence, the Q-value of the nuclearreaction is 31.848 MeV. Since the value is positive, the reaction isenergetically allowed.
Now take a 
emission nuclear reaction:
We know that:
Mass of m1 = 223.01850
m2 = 219.00948Mass of, m3 = 4.00260Q-value of this nuclear reaction isgiven as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 219.00948 − 4.00260) C2
= (0.00642 c2) u
= 0.00642 × 931.5 = 5.98 MeV
Hence, the Q value of thesecond nuclear reaction is 5.98 MeV. Since the value is positive, the reactionis energetically allowed.
Question - 26 : - Consider the fission of 
by fast neutrons. In one fission event, no neutrons are emitted and thefinal end products, after the beta decay of the primary fragments, are 
and
Calculate Q for this fission process. Therelevant atomic and particle masses are
m
=238.05079 u m
=139.90543 u
m
= 98.90594 u
Answer - 26 : - In the fission of, 10 β− particlesdecay from the parent nucleus. The nuclear reaction can be written as:
It is given that:
Mass of a nucleus
m1 = 238.05079 u
m2 = 139.90543 uMass of a nucleus, m3 = 98.90594 uMass of a neutron
m4 = 1.008665 uQ-value of the above equation,
Where,
m’ = Represents the correspondingatomic masses of the nuclei
= m1 − 92me
= m2 − 58me
= m3 − 44me
= m4
Hence, the Q-value of the fissionprocess is 231.007 MeV.
Question - 27 : - Consider the D−T reaction (deuterium−tritiumfusion)
(a) Calculate the energyreleased in MeV in this reaction from the data:
= 2.014102 u
= 3.016049 u
(b)Consider the radiusof both deuterium and tritium to be approximately 2.0 fm. What is the kineticenergy needed to overcome the coulomb repulsion between the two nuclei? To whattemperature must the gas be heated to initiate the reaction? (Hint: Kineticenergy required for one fusion event =average thermal kinetic energy availablewith the interacting particles = 2(3kT/2); k =Boltzman’s constant, T = absolute temperature.)
Answer - 27 : - (a) Take the D-T nuclear reaction 
It is given that:
Mass of
, m1= 2.014102 u
, m2 = 3.016049 uMass of m3 = 4.002603 uMass of
, m4 = 1.008665 uQ-value of the given D-T reaction is:
Q = [m1 + m2− m3 −m4] c2
= [2.014102 + 3.016049 − 4.002603 −1.008665] c2
= [0.018883 c2] u
But 1 u = 931.5 MeV/c2
∴Q = 0.018883 ×931.5 = 17.59 MeV
(b) Radius ofdeuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m
Distance between the two nuclei at the momentwhen they touch each other, d = r + r = 4 × 10−15 m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy betweenthe two nuclei is given as:
Where,
∈0 = Permittivityof free space
Hence, 5.76 × 10−14 J or 
of kinetic energy (KE) is needed to overcomethe Coulomb repulsion between the two nuclei.
However, it is given that:
KE
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kgs−2 K−1
T = Temperature required for triggering thereaction
Hence, the gas must be heated to a temperatureof 1.39 × 109 K to initiate the reaction.
Question - 28 : - Obtain the maximum kinetic energy of β-particles,and the radiation frequencies of γ decays in the decay schemeshown in Fig. 13.6. You are given that
m (198Au) = 197.968233 um (198Hg)=197.966760 u
Answer - 28 : -
It can be observed from the given γ-decaydiagram that γ1 decays from the 1.088 MeV energylevel to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decayis given as:
E1 = 1.088 − 0 =1.088 MeV
hν1= 1.088 × 1.6 × 10−19 ×106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiationradiated by γ1-decay
It can be observed from the given γ-decaydiagram that γ2 decays from the 0.412 MeV energylevel to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decayis given as:
E2 = 0.412 − 0 =0.412 MeV
hν2= 0.412 × 1.6 × 10−19 ×106 J
Where,
ν2 = Frequency of radiationradiated by γ2-decay
It can be observed from the given γ-decaydiagram that γ3 decays from the 1.088 MeV energylevel to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decayis given as:
E3 = 1.088 − 0.412= 0.676 MeV
hν3= 0.676 × 10−19 ×106 J
Where,
ν3 =Frequency of radiation radiated by γ3-decayMass of 
= 197.968233 u Mass of 
= 197.966760 u 1 u = 931.5 MeV/c2
Energy of the highest level is given as:
β1 decays from the 1.3720995MeV level to the 1.088 MeV level
∴Maximum kineticenergy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995MeV level to the 0.412 MeV level
∴Maximum kineticenergy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV