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| Chapter | 13. Nuclei |
| Subject | Physics |
| Textbook | Physics II |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
Class 12 Physics NCERT Solutions are the excellent practice set for students to ace Physics with better understandings and good marks. With the NCERT Solution for Class 12 Physics, students can confidently answer their exam questions and enjoy the best possible marks in the subject. These solutions are prepared keeping in view the latest syllabus and curriculum. In addition, the solutions are designed keeping in mind the requirements of students. Therefore, students can use these solutions for revision or to score high marks in the final examinations. Let’s get started with the solutions of Chapter 13: Nuclei.
NCERT Solutions for Class 12 Physics Chapter 13
Nuclei Solutions
Exercise
Q 13.1)
(a) Lithium has two stable isotopes 
and 
have respective abundances of 7.5% 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, 
Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of
Answer)
(a) Mass of lithium isotope , m1 = 6.01512 u
Mass of lithium isotope , m2 = 7.01600 u
Abundance of , n1= 7.5%
Abundance of , n2= 92.5%
The atomic mass of lithium atom is:
= 6.940934 u
(b) Mass of boron isotope 
m1 = 10.01294 u
Mass of boron isotope 
m2 = 11.00931 u
Abundance of , n1 = x%
Abundance of , n2= (100 − x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is:
1081.11 = 10.01294x + 1100.931 – 11.00931 x
x = 19.821/0.99637 = 19.89 %
And 100 − x = 80.11%
Hence, the abundance of is 19.89% and that of is 80.11%.
Q 13.2) The three stable isotopes of neon: 
, 
and 
have respective abundances of 90.51%, 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)
Atomic mass of , m1= 19.99 u
Abundance of , n1 = 90.51%
Atomic mass of , m2 = 20.99 u
Abundance of , n2 = 0.27%
Atomic mass of , m3 = 21.99 u
Abundance of , n3 = 9.22%
Below is the average atomic mass of neon:
= 20.1771 u
Q 13.3) Obtain the binding energy in MeV of a nitrogen nucleus 
, given m () =14.00307 u
Answer)
Atomic mass of nitrogen , m = 14.00307 u
A nucleus of nitrogen contains 7 neutrons and 7 protons.
∆m = 7mH + 7mn − m is the mass defect the nucleus
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∆m = 0.11236 × 931.5 MeV/c2
Eb = ∆mc2 is the binding energy of the nucleus
Where, c =Speed of light
= 104.66334 Mev
Therefore, 104.66334 MeV is the binding energy of the nitrogen nucleus.
Q 13.4) Obtain the binding energy of the nuclei 
and 
in units of MeV from the following data:
m () = 55.934939 u
m() = 208.980388 u
Answer)
Atomic mass of , m1 = 55.934939 u nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, ∆m = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∆m = 0.528461 × 931.5 MeV/c2
Eb1 = ∆mc2 is the binding energy of the nucleus.
Where, c = Speed of light
= 492.26 MeV
Average binding energy per nucleon = 
Atomic mass of , m2 = 208.980388 u nucleus has 83 protons and (209 − 83) 126 neutrons.
The mass defect of the nucleus is given as:
∆m’ = 83 × mH + 126 × mn − m2
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∆m’ = 1.760877 × 931.5 MeV/c2
Eb2 = ∆m’c2 is the binding energy of the nucleus.
= 1640.26 MeV
Average binding energy per nucleon = 1640.26/209 = 7.848 MeV
Q 13.5) A given coin has a mass of 3.0 g.
Calculate the nuclear energy that would be required to separate all the neutrons and protons from each
other. For simplicity assume that the coin is entirely made of 
with mass = 62.92960 u.
Answer)
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,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
atoms nucleus has 29 protons and (63 − 29) 34 neutrons
Mass defect of this nucleus, ∆m’ = 29 × mH + 34 × mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 1022
= 1.69766958 × 1022 u
But 1 u = 931.5 MeV/c2
∆m = 1.69766958 × 1022 × 931.5 MeV/c2
Eb = ∆mc2 is the binding energy of the nuclei of the coin
= 1.69766958 × 1022 × 931.5 MeV/c2 × c2
= 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10−13 J
Eb = 1.581 × 1025 × 1.6 × 10−13
= 2.5296 × 1012 J
Therefore, the energy required to separate all the neutrons and protons from the given coin is 2.5296 × 1012 J
Q 13.6) Write nuclear reaction equations for
(i) α-decay of 
(ii) α-decay of 
(iii) β−-decay of 
(iv) β– -decay of 
(v) β+-decay of 
(vi) β+ -decay of 
(vii) Electron capture of 
Answer)
In helium, α is a nucleus 
and β is an electron (e− for β− and e+ for β+). 2 protons and 4 neutrons is lost in every α decay. Whereas 1 proton and a neutrino is emitted from the nucleus in every β + decay. In every β – decay, there is a gain of 1 proton and an anti-neutrino is emitted from the nucleus.
For the given cases, the various nuclear reactions can be written as:
Q 13.7) A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?
Answer)
The half-life of the radioactive isotope = T years
N0 is the actual amount of radioactive isotope.
(a) After decay, the amount of the radioactive isotope = N
It is given that only 3.125% of N0 remains after decay. Hence, we can write:
But, 
Where, λ = Decay constant
t = Time
Hence, the isotope will take about 5T years 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 N0 remains after decay. Hence, we can write:
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.
Q 13.8) The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon.
This activity arises from the small proportion of radioactive 
present with the stable carbon isotope 
. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Answer)
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.
The decay rate of the specimen obtained from the Mohenjo-Daro site:
R’ = 9 decays/min
Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.
Therefore, the relation between the decay constant, λ and time, t is:
So, the approximate age of the Indus-Valley civilization is 4223.5 years.
Q 13.9) Obtain the amount of 
necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 
is 5.3 years.
Answer)
The strength of the radioactive source is given as:
= 8 x 10-3 x 3.7 x 1010
= 29.6 x 107 decay/s
Where,
N = Required number of atoms
Half-life of T 1/2 = 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 108 s
The rate of decay for decay constant λ is:
= 7.133 x 1016 atoms
For :
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
Mass of 7.133 x 1016 atoms = 
= 7.106 x 10-6 g
Hence, the amount of necessary for the purpose is 7.106 × 10−6 g.
Q 13.10) The half-life of 
is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer)
Half life of 
= 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m = 15 mg
90 g of atom contains 6.023 × 1023 (Avogadro’s number) atoms.
Therefore, 15 mg of contains atoms:
i.e 1.0038 x 1020 number of atoms
Rate of disintegration, 
Where,
= 7.878 x 1010 atoms/s
Hence, the disintegration rate of 15 mg of is 7.878 × 1010 atoms/s.
Q 13.11) Obtain approximately the ratio of the nuclear radii of the gold isotope 
and the 
silver isotope.
Answer)
Nuclear radius of the gold isotope = RAu
Nuclear radius of the silver isotope = RAg
Mass number of gold, AAu = 197
Mass number of silver, AAg = 107
Following is the relationship of the radii of the two nuclei and their mass number:
Hence, 1.23 is the ratio of the nuclear radii of the gold and silver isotopes.
Q 13.12) Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of
(a) 
(b) 
Given
m () = 226.02540 u, m (
) = 222.01750 u,
m () = 220.01137 u, m (
) = 216.00189 u.
Answer)
(a) Alpha particle decay of emits a helium nucleus. As a result, its mass number reduces to 222 (226 − 4) and its atomic number reduces to 86 (88 − 2). This is shown in the following nuclear reaction.
Q-value of emitted α-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
But 1 u = 931.5 MeV/c2
Q = 0.005297 × 931.5 ≈ 4.94 MeV
Kinetic energy of the α-particle = 
4.94 = 4.85 MeV
(b) Alpha particle decay of
It is given that:
Mass of = 220.01137 u
Mass of = 216.00189 u
Q-value =[ 220.01137 – ( 216.00189 + 4.00260 ) ] x 931.5
≈ 641 MeV
Kinetic energy of the α-particle = 
= 6.29 MeV
Q 13.13) The radionuclide 11C decays according to
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
m (116 C) = 11.011434 u and
m (116B ) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.
Answer)
Mass defect in the reaction is Δm = m (6C11) – m(5B11) – me
This is given in terms of atomic masses. To express in terms of nuclear mass, we should subtract 6me from carbon, 5me from boron.
Δm = m (6C11) – 6me – (m(5B11) – 5me )- me
= m (6C11) – 6me – m(5B11) + 5me – me
= m (6C11) – m(5B11) – 2 me
Δm = [11.011434 – 11.009305 – 2 x 0.000548] u
= 0.002129 – 0.001096 = 0.001033
Q = Δm x 931 MeV
= 0.001033 x 931
Q= 0.9617 MeV
The Q-factor of the reaction is equal to the maximum energy of the emitted positron.
Q 13.14) The nucleus 
decays β− by emission.
Write down the β ecay equation and determine the maximum kinetic energy of the electrons emitted.
Given that:
Answer)
In β− emission, the number of protons increases by 1, and one electron and an antineutrino is emitted from the parent nucleus.
β− emission from the nucleus 
It is given that:
Atomic mass of m () = 22.994466 u
Atomic mass of m (
) = 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
There are 10 electrons and 11 electrons in and respectively. Hence, the mass of the electron is cancelled in the Q-value equation.
Q = [ 22.994466 – 22.989770 ] c2
= (0.004696 c2) u
But 1 u = 931.5 MeV/c2
Q = 0.004696 x 931.5 = 4.374 MeV
The daughter nucleus is too heavy as compared to e– and vˉ. 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.
Q 13.15) The Q value of a nuclear reaction A + b → C + d is defined by
Q = [ mA+ mb− mC− md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
Atomic masses are given to be
Answer)
(i) The given nuclear reaction is:
It is given that:
According to the question, the Q-value of the reaction can be written as :
= [ 1.007825 + 3.016049 – 2 x 2.014102] c2
Q = ( – 0.00433 c2) u
But 1 u = 931.5 MeV/c2
Q = -0.00433 x 931.5 = – 4.0334 MeV
The negative Q-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
It is given that:
The Q-value of this reaction is given as :
= [ 2 x 12.000000 – 19.992439 – 4.002603 ] c2
= [ 0.004958 c2] u
= 0.004958 x 931.5 = 4.618377 MeV
Since we obtained positive Q-value, it can be concluded that the reaction is exothermic.
Q 13.16) Suppose, we think of fission of a 
nucleus into two equal fragments, 
. Is the fission energetically possible?
Argue by working out Q of the process. Given
Answer) The fission of can be given as :
It is given that:
Atomic mass of m () = 55.93494u
Atomic mass of m () = 27.98191u
The Q-value of this nuclear reaction is given as :
= [ 55.93494 – 2 x 27.98191 ]c2
= ( -0.02888 c2) u
But 1 u = 931.5 MeV/c2
Q = – 0.02888 x 931.5 = -26.902 MeV
Since the Q-value is negative for the fission, it is energetically not possible.
Q 13.17) The fission properties 
of are very similar to those of 
.
The average energy released per fission is 180 MeV. How much energy is released if all the atoms in 1 kg of pure undergo fission ?
Answer)
Average energy released per fission of , Eav = 180MeV
Amount of pure , m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of = 239 g
1 mole of contains NA atoms.
Therefore, mg of contains 
Total energy released during the fission of 1 kg of is calculated as:
E = Eav x 2.52 x 1024
= 180 x 2.52 x 1024
= 4.536 x 1026 MeV
Hence, 4.536 x 1026 MeV is released if all the atoms in 1 kg of pure undergo fission.
Q 13.18) A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92235 U did it contain initially?
Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92235U and that this nuclide is consumed only by the fission process.
Answer)
Reactor consumes half its fuel is 5 years. Therefore, the half-life of the fuel of the fission reactor, t1/2 = 5 x 365 x 24 x 60 x 60 s
200 MeV is released during the fission of 1 g of 92235U.
1 mole, i.e., 235 g of 92235U contains 6.023 x 1023 atoms.
Number of atoms in 1 g of 92235U is (6.023 x 1023/235) atoms.
The total energy generated per gram of 92235U is (6.023 x 1023/235) x 200 MeV/g
= 8.20 x 1010 J/g
The reactor operates only 80% of the time.
Therefore, the amount of 92235U consumed in 5 years by the 1000 MW fission reactor is
= 12614400/8.20
= 1538341 g
= 1538 Kg
Initial amount of 92235U is 2 x 1538 = 3076 Kg
Q 13.19) How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?
Take the fusion reaction as
Answer)
The fusion reaction given is
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 x 1023 atoms.
Therefore, 2 Kg of deuterium contains (6.023 x 1023/2) x 2000 = 6.023 x 1026 atoms
From the reaction given, it can be understood that 2 atoms of deuterium fuse, 3.27 MeV energy are released. The total energy per nucleus released during the fusion reaction is
E = (3.27/2) x 6.023 x 1026 MeV
= (3.27/2) x 6.023 x 1026 x 1.6 x 10 -19 x 106
= 1.576 x 1014 J
Power of the electric lamp, P = 100 W = 100 J/s
The energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is 1.576 x 1014 /100 = 1.576 x 1012s
= (1.576 x 1012)/ (365 x 24 x 60 x 60)
= (1.576 x 1012)/3.1536 x107
= 4.99 x 104 years
Q 13.20) Calculate the height of the potential barrier for a head-on collision of two deuterons.
(Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Answer)
When two deuterons collide head-on, the distance between their centers, d is given as:
Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 f m = 2 × 10−15 m
d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C
Potential energy of the two-deuteron system:
Where,
ϵ0 = Permittivity of free space 
Therefore,
= 360 k eV
Hence, the height of the potential barrier of the two-deuteron system is 360 k eV.
Q 13.21) From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
Answer)
Nuclear radius is given as
R = R0A1/3
Here,
R0 is Constant
A is the mass number of the nucleus
Nuclear matter density, ρ = Mass of the nucleus/Volume of the nucleus
Mass of the nucleus = mA
Density of the nucleus = (4/3)πR3
= (4/3)π(R0A1/3)3
= (4/3)πR03A
ρ = mA/[(4/3)πR03A]
= 3mA/(4πR03A)
ρ = 3m/(4πR03)
Nuclear matter density is nearly a constant and is independent of A.
Q 13.22) For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K−shell, is captured by the nucleus and a neutrino is emitted).
Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.
Answer)
The chemical equations of the two competing process are written as follows:
The Q-value of the positron emission reaction is determined as follows :
The Q-value of the electron capture reaction is determined as follows:
In the equation, mN is the mass of nucleus and m denotes the mass of atom.
From equation (1) and (2), we understand that,
Q1 = Q2 – 2mec2
From the above relation, we can infer that if Q1 > 0, then Q2 > 0; Also, if Q2> 0, it does not necessarily mean that Q1 > 0.
In other words, this means that if β+ 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.
Q 13.23) In a periodic table the average atomic mass of magnesium is given as 24.312 u.
The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 2412Mg (23.98504u), 2512Mg(24.98584u) and 2612Mg (25.98259u). The natural abundance of 2412Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.
Answer)
Let the abundance of 2512Mg be x%
The abundance of 2612Mg = (100 – 78.99 -x)%
= (21.01 – x)%
The average atomic mass of magnetic
24.312 x 100= 1894.57 + 24.98584x + 545. 894 – 25.98259x
2431.2 = 2440.46 – 0.99675x
0.99675x = 2440.46 – 2431.2
0.99675x = 9.26
x = 9.26/0.99675
x = 9.290%
Abundance of 2512Mg is 9.290%
The abundance of 2612Mg = (21.01 – x)%
= (21.01 -9.290 )% = 11.71%
Q 13.24) The neutron separation energy is defined as the energy required to remove a neutron from the nucleus.
Obtain the neutron separation energies of the nuclei 4120Ca and 2713Al from the following data:
m(4020Ca ) = 39.962591 u
m(4120Ca ) = 40.962278 u
m(2613Al ) = 25.986895 u
m(2713Al ) = 26.981541 u
Answer)
When the nucleons are separated from 20Ca41 → 20 Ca40 + 0n1
Mass defect, Δm = m (20 Ca40) + mn – m(20 Ca41)
= 39.962591 + 1.008665 – 40.962278
= 0.008978 amu
Neutron separation energy = 0.008978 x 931 MeV = 8.362 MeV
Similarly, 13Al27 → 13 Al26 + 0n1
Mass defect, Δm = m (13 Al26) + mn – m(13Al27)
= 25.986895 + 1.008665 – 26.981541
= 26.99556 – 26.981541 = 0.014019 amu
Neutron separation energy = 0.014019 x 931 MeV
= 13.051 MeV
Q 13.25) A source contains two phosphorous radionuclides 3215P (T1/2 = 14.3d) and 3315P(T1/2 = 25.3d).
Initially, 10% of the decays come from 3315 P. How long one must wait until 90% do so?
Answer)
Rate of disintegration is
– (dN/dt) ∝ N
Initially 10% of the decay is due to 3315 P and 90% of the decay comes from 3215P.
We have to find the time at which 90% of the decay is due to 3315 P and 10% of the decay comes from 3215P.
Initially, if the amount of 3315 P is N, then the amount of 3215P is 9N.
Finally, if the amount of 3315 P is 9N’, then the amount of 3215P is N’.
For 3215P,
On dividing, (1) by (2)
Hence, it will take about 208.5 days for 90% decay of 15P33.
Q 13.26) Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle.
Consider the following decay processes:
Calculate the Q-values for these decays and determine that both are energetically allowed.
Answer)
Δm = m (88Ra223) – m (82Pb209) – m(6C14)
= 223.01850 – 208.9817 – 14.00324
= 0.03419 u
The amount of energy released is given as
Q = Δm x 931 MeV
= 0.03419 x 931 MeV = 31.83 MeV
Δm = m (88Ra223) – m (86Rn219) – m(2He4)
= 223.01850 – 219.00948 – 4.00260
= 0.00642 u
Q = 0.00642 x 931 MeV = 5.98 MeV
As both the Q-factor values are positive the reaction is energetically allowed.
Q 13.27) Consider the fission of 23892U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 14058Ce and 9944Ru.
Calculate Q for this fission process. The relevant atomic and particle masses are
m(23892U ) =238.05079 u
m(14058Ce ) =139.90543 u
m(9944Ru ) = 98.90594 u
Answer) In the fission of 23892U, 10 β– particles decay from the parent nucleus. The nuclear reaction can be written as:
Given:
m1 = (23892U ) =238.05079 u
m2 = (14058Ce ) =139.90543 u
m3 = (9944Ru ) = 98.90594 u
m4 = 10n = 1.008665 u
The Q value is given as
Q = [ m'(23892U ) + m (10n) – m'(14058Ce ) – m'(9944Ru ) – 10me ] c2
Here,
m’ is the atomic masses of the nuclei
m'(23892U ) = m1 – 92 me
m'(14058Ce ) = m2 – 58me
m'(9944Ru ) = m3 – 44me
Therefore, Q = [m1 – 92 me+ m4 – m2 + 58me – m3 + 44me– 10me ] c2
= [m1 + m4 – m2 – m3]c2
= [238.0507 + 1.008665 – 139.90543 – 98.90594]c2
= 0.247995 c2 u[1 u = 931.5 MeV/c2]
Q = 0.247995 x 931.5 = 231.007 MeV
The Q- value of the fission process is 231.007 MeV
Q 13.28) Consider the D–T reaction (deuterium–tritium fusion)
(a) Calculate the energy released in MeV in this reaction from the data:
m(21H )=2.014102 u
m(31H ) =3.016049 u
(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
Answer)
The Q value is given as
Q = Δm x 931 MeV
= (m (1H2) + m(1H3) – m(2He4) – mn) x 931
=( 2.014102 + 3.016049- 4.002603-1.00867) x 931
Q = 0.0188 x 931 = 17.58 MeV
(b) Repulsive potential energy of two nuclei when they touch each other is
Where ε0 = Permittivity of free space
= (23.04 x 10-29)/(4 x 10-15)
= 5. 76 x 10-14 J
Kinetic energy needed to overcome the coulomb repulsion between the two nuclei is
K.E = 5. 76 x 10-14 J
K.E= 2 x (3/2) KT
K is the Boltzmann constant = 1.38 x 10-23
T = K.E/3K = 5. 76 x 10-14/(3 x 1.38 x 10-23)
= 5. 76 x 10-14/(4.14 x 10-23)
= 1.3913 x 109 K
Q 13.29) Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that
m(198Au) = 197.968233 u
m(198Hg) =197.966760 u
Answer)
From the γ decay diagram it can be seen that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.
Energy corresponding to γ1 decay is given as
E1 =1.088−0=1.088MeV
hv1 =1.088×1.6×10−19 ×106J
Where,
h= Planck’s constant = 6.6 x 10-34 Js
v1 = frequency of radiation radiated by γ1 decay.
From the diagram it can observed that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.
Energy corresponding to γ2 decay is given as
E2 =0.412−0=0.412MeV
hv2 =0.412×1.6×10−19 × 106 J
Where v2 is the frequency of the radiation radiated by γ2 decay.
Therefore, v2 =E2/h
=9.988×1019 Hz
From the gamma decay diagram, it can be γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Energy corresponding to γ3 is given as
E3 =1.088−0.412=0.676 MeV
hv3 =0.676×10−19 ×106 J
Where v3 = frequency of the radiation radiated by γ3 decay
Therefore, v3 =E3/h
=1.639×1020 Hz
The mass of Au is 197.968233 u
Mass of Hg = 197.966760 u[1u=931.5MeV/c2]
The energy of the highest level is given as :
E = [m(19878Au−m(19080Hg)]
= 197.968233 − 197.96676=0.001473u
= 0.001473×931.5 = 1.3720995MeV
β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
Therefore, the maximum kinetic energy of the β1 particle = 1.3720995−1.088 = 0.2840995MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
Therefore, the maximum kinetic energy of the β2 particle = 1.3720995−0.412 = 0.9600995
Q 13.30) Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within the sun and b) the fission of 1.0 kg of 235U in a fission reactor.
Answer)
Amount of hydrogen, m = 1 kg = 1000 g
(a) 1 mole, i.e., 1 g of hydrogen contains 6.023 x 1023 atoms. Therefore, 1000 g of hydrogen contains 6.023 x 1023 x 1000 atoms.
In Sun, four hydrogen nuclei fuse to form a helium nucleus and will release energy of 26 MeV
Energy released by the fusion of 1 kg of hydrogen,
E1 = (6.023 x 1023 x 1000 x 26)/4
E1= 39.16 x 1026 MeV
(b) Amount of 235U = 1 kg = 1000 g
1 mole, i.e., 235 g of uranium contains 6.023 x 1023 atoms. Therefore, 1000 g of uranium contains (6.023 x 1023 x 1000)/235 atoms.
The energy released during the fission of one atom of 235U = 200 MeV
Energy released by the fission of 1 Kg of 235U,
E2= (6.023 x 1023 x 1000 x 200)/235
E2 = 5.1 x 1026 MeV
(E1/E2) = (39.16 x 1026/5.1 x 1026)
= 7.67
The energy released in the fusion of hydrogen is 7.65 times more than the energy released during the fission of Uranium.
Q 13.31) Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten per cent of which was to be obtained from nuclear power plants.
Suppose we are given that, on average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.
Answer)
Electric power to be generated = 2 x 105 MW = 2 x 105 x 106 J/s = 2 x 1011 J/s
10 % of the amount is obtained from the nuclear power plant
P1 = (10/100) x 2 x 1011 x 60 x 60 x 24 x 365 J/year
Heat energy released during per fission of a 235U nucleus, E = 200 MeV
Efficiency of the reactor = 25%
The amount of energy converted to electrical energy in fission is (25/100) x 200 = 50 MeV
= 50 x 1.6 x 10-19 x 106 = 80 x 10-13 J
Required number of atoms for fission per year
= [(10/100) x 2 x 1011 x 60 x 60 x 24 x 365]/80 x 10-13
= 788400 x 1023 atoms
1 mole, i.e., 235 g of 235U contains 6.023 x 1023 atoms
Mass of 6.023 x 1023 atoms of 235U = 235 x 10-3 kg
Mass of 788400 x 1023 atoms of 235U
=[ (235 x 10-3)/(6.023 x 1023) ] x 78840 x 1024
= 3. 076 x 104 Kg
The mass of uranium needed per year is 3.076 x 104 Kg
That’s it. These were the solutions of NCERT Class 12 Physics Chapter 13 – Nuclei. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.