Khopdi Kharab: Nuclei

Ans: 1
(a) Mass of lithium isotope

,m₁= 6.01512 u
Mass of lithium isotope

,m₂= 7.01600 u
Abundance of

,η₁= 7.5%
Abundance of

,η₂= 92.5%
The atomic mass of lithium atom is given as:

(b) Mass of boron isotope

,m₁= 10.01294 u
Mass of boron isotope

,m₂= 11.00931 u
Abundance of

,η₁= x%
Abundance of

,η₂= (100 − x)%
Atomic mass of boron, m= 10.811 u
Theatomic mass of boron atom is given as:

And100 − x= 80.11%
Hence,the abundance of

is 19.89% and that of

is 80.11%.
Ans: 2
Atomic mass of

,m₁= 19.99 u
Abundanceof

,η₁= 90.51%
Atomic mass of

,m₂= 20.99 u
Abundance of

,η₂= 0.27%
Atomic mass of

,m₃= 21.99 u
Abundance of

,η₃= 9.22%
Theaverage atomic mass of neon is given as:

Ans: 3
Atomic mass of nitrogen

,m = 14.00307 u
Anucleus of nitrogen

contains 7 protons and 7 neutrons.
Hence,the mass defect of this nucleus, Δm= 7m_H+ 7m_n− m
Where,
Mass of a proton, m_H= 1.007825 u
Mass of a neutron, m_n= 1.008665 u
∴Δm= 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But1 u = 931.5 MeV/c²
∴Δm= 0.11236 × 931.5 MeV/c²
Hence,the binding energy of the nucleus is given as:
E_b= Δmc²
Where,
c= Speed of light
∴E_b= 0.11236 ×931.5

= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

Ans: 4
Atomicmass of

,m₁= 55.934939 u

nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence,the mass defect of the nucleus, Δm= 26 × m_H+ 30 × m_n− m₁
Where,
Mass of a proton, m_H= 1.007825 u
Mass of a neutron, m_n= 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 this nucleus is given as:
E_b₁= Δmc²
Where,
c= Speed of light
∴E_b₁= 0.528461 × 931.5

= 492.26 MeV
Average binding energy per nucleon

Atomic mass of

,m₂= 208.980388 u

nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence,the mass defect of this nucleus is given as:
Δm' = 83 × m_H+ 126 × m_n− m₂
Where,
Mass of a proton, m_H= 1.007825 u
Mass of a neutron, m_n= 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 this nucleus is given as:
E_b₂= Δm'c²
= 1.760877 × 931.5

= 1640.26 MeV
Average bindingenergy per nucleon =

Ans: 5
Mass of a copper coin, m’= 3 g
Atomic mass of

atom, m = 62.92960 u
The total number of

atoms in the coin

Where,
N_A= Avogadro’s number = 6.023 × 10²³atoms /g
Mass number = 63 g

nucleus has 29 protons and (63 − 29) 34 neutrons
∴Mass defect of this nucleus, Δm' = 29 × m_H+ 34 × m_n− m
Where,
Mass of a proton, m_H= 1.007825 u
Mass of a neutron, m_n= 1.008665 u
∴Δm' = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, Δm= 0.591935 × 2.868 × 10²²
= 1.69766958 × 10²²u
But 1 u = 931.5 MeV/c²
∴Δm= 1.69766958 × 10²²× 931.5 MeV/c²
Hence, the binding energy of the nuclei of the coin is given as:
E_b=Δmc²
= 1.69766958 × 10²²× 931.5

= 1.581 × 10²⁵MeV
But 1 MeV = 1.6 × 10⁻¹³J
E_b= 1.581 × 10²⁵× 1.6 × 10⁻¹³
= 2.5296 × 10¹²J
Thismuch energy is required to separate all the neutrons and protons from the given coin.

Ans: 6
α is a nucleus of helium

and βis an electron (e⁻for β⁻and e⁺for β⁺).In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β⁺-decay, there is a loss of 1 proton and a neutrino 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, the various nuclear reactions can be written as:

Ans: 7
Half-life of the radioactive isotope = Tyears
Original amount of the radioactive isotope = N₀
(a) After decay, the amount of the radioactive isotope = N
It is given that only 3.125% of N₀remains after decay. Hence, we can write:

Where,
λ= Decay constant
t= Time

Hence, the isotope will take about 5Tyears to reduce to 3.125% of its original value.
(b) After decay, the amount of the radioactive isotope = N
It is given that only 1% of N₀remains after decay. Hence, we can write:

Since,λ= 0.693/T

Hence, the isotope will take about 6.645Tyears to reduce to 1% of its original value.

Ans: 8
Decay rate of living carbon-containing matter, R= 15 decay/min
Let N be the number of radioactive atoms present in a normal carbon-containing matter.
Half life of

= 5730 years
The decay rate of the specimen obtained from the Mohenjodaro site:
R' = 9 decays/min
Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate the decay constant, λand time, tas:

Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.

Ans: 9
The strength of the radioactive source is given as:

Where,
N= Required number of atoms
Half-life of

= 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 10⁸s
For decay constant λ,we have the rate of decay as:

Where,λ

For

:
Mass of 6.023 × 10²³(Avogadro’s number) atoms = 60 g
∴Mass of

atoms

Hence,the amount of

necessary for the purpose is 7.106 × 10⁻⁶g.

Ans: 10
Half life of

= 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 10⁸s
Mass of the isotope, m= 15 mg
90 g of

atom contains 6.023 × 10²³(Avogadro’s number) atoms.
Therefore,15 mg of

contains:

Rate of disintegration,

Where,
λ =Decay constant

Hence, the disintegration rate of 15 mg of the given isotope is
7.878 × 10¹⁰atoms/s.

Ans: 11
Nuclear radius of the gold isotope

= R_Au
Nuclearradius of the silver isotope

= R_Ag
Mass number of gold, A_Au= 197
Mass number of silver, A_Ag= 107
The ratio of the radii of the two nuclei is related with their mass numbers as:

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Ans: 12
(a) Alpha particle decay of

emits a helium nucleus. As a result, its mass number reduces to (226 −4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

Q-value of
emittedα-particle = (Sum of initial mass − Sum of final mass) c²
Where,
c= Speed of light
It is given that:

Q-value = [226.02540 − (222.01750 + 4.002603)] u c²
= 0.005297 u c²
But1 u = 931.5 MeV/c²
∴Q= 0.005297 × 931.5 ≈4.94 MeV
Kinetic energy of the α-particle

(b) Alpha particle decay of

is shown by the following nuclear reaction.

It is given that:
Mass of

= 220.01137 u
Mass of

= 216.00189 u
∴Q-value =

≈ 641 MeV
Kinetic energy of the α-particle

= 6.29 MeV

Ans: 13
The given nuclear reaction is:

Atomic mass of

= 11.011434 u
Atomic mass of

Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the

nucleus is given as:

Where,
m_e= Mass of an electron or positron = 0.000548 u
c= Speed of light
m’= Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 m_ein the case of

and 5 m_ein the case of

.
Hence, equation (1) reduces to:

∴ΔQ= [11.011434 − 11.009305 − 2 ×0.000548] c²
= (0.001033 c²) u
But 1 u = 931.5 Mev/c²
∴ΔQ= 0.001033 ×931.5 ≈0.962 MeV
The value of Qis almost comparable to the maximum energy of the emitted positron.

Ans: 14
In

emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

emission of the nucleus

is given as:

It is given that:
Atomic mass of

= 22.994466 u
Atomic mass of

= 22.989770 u
Mass of an electron, m_e= 0.000548 u
Q-value of the given reaction is given as:

There are 10 electrons in

and 11 electrons in

. Hence, the mass of the electron is cancelled in the Q-value equation.

The daughter nucleus is too heavy as compared to

and

. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

Ans: 15
(i) The given nuclear reaction is:

It is given that:
Atomic mass

Atomic mass

According to the question, the Q-value of the reaction can be written as:

The negativeQ-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:

It is given that:
Atomic mass of

Atomic mass of

TheQ-value of this reaction is given as:

The positive Q-value of the reaction shows that the reaction is exothermic.

Ans: 16
The fission of

can be given as:

It is given that:
Atomic mass of

= 55.93494 u
Atomic mass of

TheQ-value of this nuclear reaction is given as:

TheQ-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, theQ-value must be positive.

Ans: 17
Average energy released per fission of

Amount of pure

,m = 1 kg = 1000 g
N_A= Avogadro number = 6.023 ×10²³
Mass number of

= 239 g
1 mole of

contains N_Aatoms.
∴mg of

contains

∴Total energy released during the fission of 1 kg of

is calculated as:

Hence,

is released if all the atoms in 1 kg of pure

undergo fission.

Ans: 18
Half life of the fuel of the fission reactor,

years
= 5 × 365 × 24 × 60 × 60 s
We know that in the fission of 1 g of

nucleus, the energy released is equal to 200 MeV.
1 mole, i.e., 235 g of

contains 6.023 ×10²³atoms.
∴1 g

contains

The total energy generated per gram of

is calculated as:

The reactor operates only 80% of the time.
Hence, the amount of

consumed in 5 years by the 1000 MW fission reactor is calculated as:

∴Initial amount of

= 2 ×1538 = 3076 kg

Ans: 19
The given fusion reaction is:

Amount of deuterium, m= 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 ×10²³atoms.
∴2.0 kg of deuterium contains

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.
∴Total energy per nucleus released in the fusion reaction:

Power of the electric lamp, P= 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:

Ans: 20
When two deuterons collide head-on, the distance between their centres,d is given as:
Radius of 1^stdeuteron + Radius of 2^nddeuteron
Radius of a deuteron nucleus = 2 fm = 2 ×10⁻¹⁵m
∴d= 2 ×10⁻¹⁵+ 2 ×10⁻¹⁵= 4 ×10⁻¹⁵m
Charge on a deuteron nucleus = Charge on an electron = e= 1.6 ×10⁻¹⁹C
Potential energy of the two-deuteron system:

Where,

= Permittivity of free space

Hence, the height of the potential barrier of the two-deuteron system is
360 keV.

Ans: 21
We have the expression for nuclear radius as:
R= R₀A¹/³
Where,
R₀= Constant.
A= Mass number of the nucleus
Nuclear matter density,

Letm be the average mass of the nucleus.
Hence, mass of the nucleus = mA

Hence, the nuclear matter density is independent of A. It is nearly constant.

Ans: 22
Let the amount of energy released during the electron capture process beQ₁. The nuclear reaction can be written as:

Let the amount of energy released during the positron capture process beQ₂. The nuclear reaction can be written as:

= Nuclear mass of

= Atomic mass of

m_e= Mass of an electron
c= Speed of light
Q-value of the electron capture reaction is given as:

Q-value of the positron capture reaction 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, this means that if

emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

Ans: 23
Average atomic mass of magnesium, m= 24.312 u
Mass of magnesium isotope

,m₁= 23.98504 u
Mass of magnesium isotope

,m₂= 24.98584 u
Mass of magnesium isotope

,m₃= 25.98259 u
Abundance 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 atomic mass as:

Hence, the abundance of

is 9.3% and that of

is 11.71%.

Ans: 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 u
Mass

) = 40.962278 u
Mass

= 1.008665 u
The mass defect of this reaction is given as:
Δm=

∴Δm= 0.008978 × 931.5 MeV/c²
Hence, the energy required for neutron removal is calculated as:

For

, the neutron removal reaction can be written as:

It is given that:
Mass

= 26.981541 u
Mass

= 25.986895 u
The mass defect of this reaction is given as:

Hence, the energy required for neutron removal is calculated as:

Ans: 25
Half life of

,T_{1/2 =} 14.3 days
Half life of

,T’_1/2= 25.3 days

nucleus decay is 10% of the total amount of decay.
The source has initially 10% of

nucleus and 90% of

nucleus.
Suppose after tdays, the source has 10% of

nucleus and 90% of

nucleus.
Initially:
Number of

nucleus = N
Number of

nucleus = 9 N
Finally:
Number of

Number of

For

nucleus, we can write the number ratio as:

For

, we can write the number ratio as:

On dividing equation (1) by equation (2), we get:

Hence, it will take about 208.5 days for 90% decay of

.

Ans: 26
Take a

emission nuclear reaction:

We know that:
Mass of

m₁= 223.01850 u
Mass of

m₂= 208.98107 u
Mass of

,m₃= 14.00324 u
Hence, the Q-value of the reaction 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 nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.
Now take a

emission nuclear reaction:

We know that:
Mass of

m₁= 223.01850
Mass of

m₂= 219.00948
Mass of

,m₃= 4.00260
Q-value of this nuclear reaction is given 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 Qvalue of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

Ans: 27
In the fission of

, 10 β⁻particles decay from the parent nucleus. The nuclear reaction can be written as:

It is given that:
Mass of a nucleus

m₁= 238.05079 u
Mass of a nucleus

m₂= 139.90543 u
Mass of a nucleus

,m₃ = 98.90594 u
Mass of a neutron

m₄= 1.008665 u
Q-value of the above equation,

Where,
m’ = Represents the corresponding atomic masses of the nuclei

=m₁ − 92m_e

=m₂ − 58m_e

=m₃ − 44m_e
=m₄

Hence, the Q-value of the fission process is 231.007 MeV.

Ans: 28
(a) Take the D-T nuclear reaction:
It is given that:
Mass of

,m₁= 2.014102 u
Mass of

,m₂= 3.016049 u
Mass of

m₃= 4.002603 u
Mass of

,m₄= 1.008665 u
Q-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 of deuterium and tritium, r≈2.0 fm = 2 × 10⁻¹⁵m
Distance between the two nuclei at the moment when they touch each other, d = r + r = 4 × 10⁻¹⁵m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy between the two nuclei is given as:

Where,
∈₀= Permittivity of free space

Hence, 5.76 × 10⁻¹⁴J or

of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.
However,it is given that:
KE

Where,
k= Boltzmann constant = 1.38 × 10⁻²³m² kg s⁻²K⁻¹
T = Temperature required for triggering the reaction

Hence, the gas must be heated to a temperature of 1.39 × 10⁹K to initiate the reaction.

Ans: 29
It can be observed from the given γ-decay diagram that γ₁decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ₁-decay is given as:
E₁= 1.088 − 0 = 1.088 MeV
hν₁= 1.088 × 1.6 × 10⁻¹⁹× 10⁶J
Where,
h= Planck’s constant = 6.6 × 10⁻³⁴Js
ν₁= Frequency of radiation radiated byγ₁-decay

It can be observed from the given γ-decay diagram that γ₂decays from the 0.412 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ₂-decay is given as:
E₂= 0.412 − 0 = 0.412 MeV
hν₂= 0.412 × 1.6 × 10⁻¹⁹× 10⁶J
Where,
ν₂= Frequency of radiation radiated byγ₂-decay

It can be observed from the given γ-decay diagram that γ₃decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ₃-decay is given as:
E₃= 1.088 − 0.412 = 0.676 MeV
hν₃= 0.676 × 10⁻¹⁹× 10⁶J
Where,
ν₃= Frequency of radiation radiated byγ₃-decay

Mass of

= 197.968233 u
Mass of

= 197.966760 u
1 u = 931.5 MeV/c²
Energy of the highest level is given as:

β₁decays from the 1.3720995 MeV level to the 1.088 MeV level
∴Maximum kinetic energy of the β₁particle = 1.3720995 − 1.088
= 0.2840995 MeV
β₂decays from the 1.3720995 MeV level to the 0.412 MeV level
∴Maximum kinetic energy of the β₂particle = 1.3720995 − 0.412
= 0.9600995 MeV

Ans: 30
(a) Amount of hydrogen, m= 1 kg = 1000 g
1 mole, i.e., 1 g of hydrogen (

) contains 6.023 × 10²³atoms.
∴1000 g of

contains 6.023 × 10²³× 1000 atoms.
Within the sun, four

nuclei combine and form one

nucleus. In this process 26 MeV of energy is released.
Hence,the energy released from the fusion of 1 kg

is:

(b) Amount of

= 1 kg = 1000 g
1mole, i.e., 235 g of

contains 6.023 × 10²³atoms.
∴1000 g of

contains

It is known that the amount of energy released in the fission of one atom of

is 200 MeV.
Hence, energy released from the fission of 1 kg of

is:

∴

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium.

Ans: 31
Amount of electric power to be generated, P= 2 × 10⁵MW
10% of this amount has to be obtained from nuclear power plants.
∴Amount of nuclear power,

= 2 × 10⁴MW
= 2 × 10⁴× 10⁶J/s
= 2 × 10¹⁰× 60 × 60 × 24 × 365 J/y
Heat energy released per fission of a ²³⁵U nucleus, E= 200 MeV
Efficiency of a reactor = 25%
Hence, the amount of energy converted into the electrical energy per fission is calculated as: