Chapter 13 Nuclei Solutions
Question - 1 : - (a) Two stable isotopes of lithium 
and 
have respective abundances of7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u,respectively. Find the atomic mass of lithium.
(b)Boron has two stable isotopes, 
and
. Their respective masses are 10.01294 u and 11.00931u, and the atomic mass of boron is 10.811 u. Find the abundances of and
Answer - 1 : - (a) Massof lithium isotope , m1 =6.01512 u
Massof lithium isotope , m2 =7.01600 u
Abundanceof , η1= 7.5%
Abundanceof , η2=92.5%
The atomic mass of lithium atomis given as:
(b) Massof boron isotope , m1 =10.01294 u
Massof boron isotope , m2 =11.00931 uAbundance of , η1 = x%
Abundanceof , η2= (100− x)%Atomic mass of boron, m =10.811 u
The atomic mass of boronatom is given as:
And 100 − x =80.11%
Hence, theabundance of is 19.89% and that of is 80.11%.
Question - 2 : - The three stable isotopes of neon: 
and 
have respective abundances of90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u,20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer - 2 : - Atomic mass of 
, m1=19.99 u
Abundance of , η1 =90.51%
Atomicmass of 
, m2 =20.99 u
Abundanceof , η2 =0.27%
Atomicmass of , m3 =21.99 u
Abundanceof , η3 =9.22%
The average atomic mass ofneon is given as:
Question - 3 : - Obtain the binding energy (in MeV) of anitrogen nucleus
, given 
=14.00307 u
Answer - 3 : - Atomic mass of nitrogen
, m = 14.00307 u
A nucleusof nitrogen 
contains 7 protons and 7neutrons.
Hence, the mass defect ofthis nucleus, Δm = 7mH + 7mn − m
Where,
Mass of a proton, mH =1.007825 u
Mass of a neutron, mn=1.008665 u
∴Δm =7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∴Δm =0.11236 × 931.5 MeV/c2
Hence, the binding energy ofthe nucleus is given as:
Eb = Δmc2
Where,
c =Speed of light
∴Eb =0.11236 × 931.5 
= 104.66334 MeV
Hence, the binding energy of anitrogen nucleus is 104.66334 MeV.
and 
in units of MeV from thefollowing data:
= 55.934939 u 
= 208.980388 u
Answer - 4 : - Atomic mass of , m1 =55.934939 u
nucleus has 26 protons and(56 − 26) = 30 neutrons Hence, the mass defect ofthe nucleus, Δm = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH =1.007825 u
Mass of a neutron, mn =1.008665 u
∴Δm =26 × 1.007825 + 30 × 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∴Δm =0.528461 × 931.5 MeV/c2
The binding energy of thisnucleus is given as:
Eb1 = Δmc2
Where,
c =Speed of light
∴Eb1 =0.528461 × 931.5 = 492.26 MeV
Averagebinding energy per nucleon 
Atomic mass of, m2 =208.980388 u nucleus has 83 protons and(209 − 83) 126 neutrons. Hence, the mass defect ofthis nucleus is given as:
Δm' = 83 × mH +126 × mn − m2
Where,
Mass of a proton, mH =1.007825 u
Mass of a neutron, mn =1.008665 u
∴Δm'= 83 × 1.007825 + 126 × 1.008665 − 208.980388
= 83.649475 + 127.091790 −208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∴Δm'= 1.760877 × 931.5 MeV/c2
Hence, the binding energy of thisnucleus is given as:
Eb2 = Δm'c2
=1.760877 × 931.5
= 1640.26 MeV
Averagebindingenergy per nucleon = 
atoms (of mass 62.92960 u).
Answer - 5 : -
Mass of a copper coin, m’= 3 g
Atomicmass of
atom, m =62.92960 uThetotal number of atoms in the coin
Where,
NA = Avogadro’snumber = 6.023 × 1023 atoms /g
Mass number = 63 g
nucleus has 29 protons and (63 −29) 34 neutrons
∴Massdefect of this nucleus, Δm' = 29 × mH + 34× mn − m
Where,
Mass of a proton, mH =1.007825 u
Mass of a neutron, mn =1.008665 u
∴Δm'= 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all theatoms present in the coin, Δm = 0.591935 × 2.868 × 1022
= 1.69766958 × 1022 u
But 1 u = 931.5 MeV/c2
∴Δm =1.69766958 × 1022 × 931.5 MeV/c2
Hence, the binding energy of thenuclei of the coin is given as:
Eb= Δmc2
=1.69766958 × 1022 × 931.5 = 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10−13 J
Eb = 1.581 ×1025 × 1.6 × 10−13
= 2.5296 × 1012 J
This much energy is requiredto separate all the neutrons and protons from the given coin.
Write nuclear reaction equationsfor
(i) α-decayof 
(ii) α-decayof 
(iii) β−-decayof 
(iv) β−-decayof 
(v) β+-decayof 
(vi) β+-decayof 
(vii) Electroncapture of 
Answer - 6 : - α is a nucleus of helium 
and β is anelectron (e− for β− and e+ for β+).In every α-decay, there is a loss of 2 protons and 4 neutrons. Inevery β+-decay, there is a loss of 1 proton and aneutrino is emitted from the nucleus. In every β−-decay,there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
For the given cases, thevarious nuclear reactions can be written as:
Question - 7 : - A radioactive isotope has ahalf-life of T years. How long will it take the activity toreduce to a) 3.125%, b) 1% of its original value?
Answer - 7 : -
Half-life of the radioactiveisotope = T years
Original amount of theradioactive isotope = N0
(a) Afterdecay, the amount of the radioactive isotope = N
It is given that only 3.125%of N0 remains after decay. Hence, we can write:
Where,
λ = Decay constant
t =Time
Hence, the isotope will takeabout 5T years to reduce to 3.125% of its original value.
(b) Afterdecay, the amount of the radioactive isotope = N
It is given that only 1% of N0 remainsafter decay. Hence, we can write:
Since, λ =0.693/T
Hence, the isotope will takeabout 6.645T years to reduce to 1% of its original value.
Question - 8 : - The normal activity of living carbon-containing matteris found to be about 15 decays per minute for every gram of carbon. Thisactivity arises from the small proportion of radioactive 
present with the stable carbon isotope
. When the organism is dead, its interactionwith the atmosphere (which maintains the above equilibrium activity) ceases andits activity begins to drop. From the known half-life (5730 years) of, and the measured activity, the age of the specimencan be approximately estimated. This is the principle of dating used in archaeology.Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minuteper gram of carbon. Estimate the approximate age of the Indus-Valleycivilisation.
Answer - 8 : -
Decay rate of livingcarbon-containing matter, R = 15 decay/min
Let N be the number ofradioactive atoms present in a normal carbon- containing matter.
Halflife of, 
= 5730 yearsThe decay rate of the specimenobtained from the Mohenjodaro site:
R' =9 decays/min
Let N' be the number ofradioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate thedecay constant, λand time, t as:
Hence, the approximate age of theIndus-Valley civilisation is 4223.5 years.
Question - 9 : - Obtain the amount of 
necessary to provide a radioactive source of 8.0 mCistrength. The half-life of is 5.3 years.
Answer - 9 : -
The strength of the radioactivesource is given as:
Where,
N =Required number of atoms
Half-lifeof, 
= 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 108 s
For decay constant λ,we have the rate of decay as:
Where, λ 
For
Mass of 6.023 × 1023 (Avogadro’snumber) atoms = 60 g
∴Mass of 
atoms 
Hence, the amount of necessary for the purposeis 7.106 × 10−6 g.
Question - 10 : - The half-life of 
is 28 years. What is thedisintegration rate of 15 mg of this isotope?
Answer - 10 : - Half life of , 
= 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m =15 mg
90g of atom contains 6.023 × 1023 (Avogadro’snumber) atoms.Therefore, 15mg of contains:
Rate of disintegration, Where,
λ =Decay constant 
Hence, the disintegration rate of15 mg of the given isotope is 7.878 × 1010 atoms/s.