NCERT Solutions - Nuclei Class 12 Notes | EduRev

Physics Class 12

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Class 12 : NCERT Solutions - Nuclei Class 12 Notes | EduRev

The document NCERT Solutions - Nuclei Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
All you need of Class 12 at this link: Class 12

NCERT QUESTION

(Nuclei)
Ques 13.1:
(a) Two stable isotopes of lithium NCERT Solutions - Nuclei Class 12 Notes | EduRev  and NCERT Solutions - Nuclei Class 12 Notes | EduRev 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, NCERT Solutions - Nuclei Class 12 Notes | EduRev and NCERT Solutions - Nuclei Class 12 Notes | EduRev. 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev  and NCERT Solutions - Nuclei Class 12 Notes | EduRev.
Ans: (a) Mass of lithium isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev , m1 = 6.01512 u
Mass of lithium isotope  NCERT Solutions - Nuclei Class 12 Notes | EduRev,  m2 = 7.01600 u
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η1= 7.5%
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η2= 92.5%
The atomic mass of lithium atom is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
(b) Mass of boron isotope  NCERT Solutions - Nuclei Class 12 Notes | EduRev, m1 = 10.01294 u
Mass of boron isotope  NCERT Solutions - Nuclei Class 12 Notes | EduRev, m2 = 11.00931 u
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η1 = x%
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η2= (100 − x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
And 100 − x = 80.11%
Hence, the abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev is 19.89% and that of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 80.11%.


Ques 13.2:
The three stable isotopes of neon:  NCERT Solutions - Nuclei Class 12 Notes | EduRevand NCERT Solutions - Nuclei Class 12 Notes | EduRev 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.
Ans: Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, m1= 19.99 u
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η1 = 90.51%
Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, m2 = 20.99 u
Abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev , η2 = 0.27%
Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, m3 = 21.99 u
Abundance of  NCERT Solutions - Nuclei Class 12 Notes | EduRev, η3 = 9.22%
The average atomic mass of neon is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
  = 20.1771 u


Ques 13.3:
Obtain the binding energy (in MeV) of a nitrogen nucleus NCERT Solutions - Nuclei Class 12 Notes | EduRev, given NCERT Solutions - Nuclei Class 12 Notes | EduRev
=14.00307 u
Ans: Atomic mass of nitrogen NCERT Solutions - Nuclei Class 12 Notes | EduRev, m = 14.00307 u
A nucleus of nitrogen NCERT Solutions - Nuclei Class 12 Notes | EduRev  contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7mH  7mnm
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 of the nucleus is given as:
Eb = Δmc2
Where,
c = Speed of light
Eb = 0.11236 × 931.5 NCERT Solutions - Nuclei Class 12 Notes | EduRev
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.


Ques 13.4:
Obtain the binding energy of the nuclei NCERT Solutions - Nuclei Class 12 Notes | EduRev  and NCERT Solutions - Nuclei Class 12 Notes | EduRev in units of MeV from the following data:
NCERT Solutions - Nuclei Class 12 Notes | EduRev = 55.934939 u , NCERT Solutions - Nuclei Class 12 Notes | EduRev= 208.980388 u
Ans: Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m1 = 55.934939 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, Δm = 26 × mH  30 × mnm1
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 this nucleus is given as:
Eb1 = Δmc2
Where,
c = Speed of light
Eb1 = 0.528461 × 931.5 NCERT Solutions - Nuclei Class 12 Notes | EduRev
= 492.26 MeV
Average binding energy per nucleon NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m2 = 208.980388 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 × mH  126 × mnm2
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 this nucleus is given as:
Eb2 = Δm'c2
= 1.760877 × 931.5NCERT Solutions - Nuclei Class 12 Notes | EduRev
= 1640.26 MeV
Average binding energy per nucleon = NCERT Solutions - Nuclei Class 12 Notes | EduRev


Ques 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev atoms (of mass 62.92960 u).
Ans: Mass of a copper coin, m’ = 3 g
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev atom, m = 62.92960 u
The total number of NCERT Solutions - Nuclei Class 12 Notes | EduRev atoms in the coinNCERT Solutions - Nuclei Class 12 Notes | EduRev
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRevnucleus has 29 protons and (63 − 29) 34 neutrons
∴ Mass defect of this nucleus, Δm' = 29 × mH  34 × mnm
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 the atoms 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 the nuclei of the coin is given as:
Eb= Δmc2
= 1.69766958 × 1022 × 931.5 NCERT Solutions - Nuclei Class 12 Notes | EduRev
= 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 required to separate all the neutrons and protons from the given coin.


Ques 13.6:
Write nuclear reaction equations for
(i) α-decay of  NCERT Solutions - Nuclei Class 12 Notes | EduRev
(ii) α-decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev
(iii) β-decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev
(iv) β-decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev
(v) β -decay of  NCERT Solutions - Nuclei Class 12 Notes | EduRev
(vi) β -decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev
(vii) Electron capture of NCERT Solutions - Nuclei Class 12 Notes | EduRev
Ans: α is a nucleus of helium NCERT Solutions - Nuclei Class 12 Notes | EduRev and β is 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 anti-neutrino is emitted from the nucleus.

For the given cases, the various nuclear reactions can be written as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev

Ques 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?
Ans: Half-life of the radioactive isotope = T years
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 N0 remains after decay. Hence, we can write:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Where,
λ = Decay constant
t = Time
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev

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:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Since, λ = 0.693/T

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

Ques 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev  present with the stable carbon isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev . 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of NCERT Solutions - Nuclei Class 12 Notes | EduRev 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 civilization.
Ans: 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev,  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 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
ohenjodaro period.
Therefore, we can relate the decay constant, λand time, t as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.

Ques 13.9:
Obtain the amount of NCERT Solutions - Nuclei Class 12 Notes | EduRev necessary to provide a radioactive source of 8.0 mCi strength. The half-life of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 5.3 years.
Ans: The strength of the radioactive source is given as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Where,
N = Required number of atoms
Half-life of NCERT Solutions - Nuclei Class 12 Notes | EduRev, NCERT Solutions - Nuclei Class 12 Notes | EduRev  = 5.3 years
= 5.3 × 365 × 24 × 60 × 60
= 1.67 × 108 s
For decay constant λ, we have the rate of decay as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

Where, λ NCERT Solutions - Nuclei Class 12 Notes | EduRev

NCERT Solutions - Nuclei Class 12 Notes | EduRev

NCERT Solutions - Nuclei Class 12 Notes | EduRev

For NCERT Solutions - Nuclei Class 12 Notes | EduRev:
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
∴Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev atoms NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the amount of NCERT Solutions - Nuclei Class 12 Notes | EduRev  necessary for the purpose is 7.106 × 10−6 g.

Ques 13.10:
The half-life of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 28 years. What is the disintegration rate of 15 mg of this isotope?
Ans: Half life of NCERT Solutions - Nuclei Class 12 Notes | EduRev , NCERT Solutions - Nuclei Class 12 Notes | EduRev = 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m = 15 mg
90 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev atom contains 6.023 × 1023 (Avogadro’s number) atoms.
Therefore, 15 mg of  NCERT Solutions - Nuclei Class 12 Notes | EduRev contains:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

Rate of disintegration, NCERT Solutions - Nuclei Class 12 Notes | EduRev

Where,

λ = Decay constant NCERT Solutions - Nuclei Class 12 Notes | EduRev

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

Ques 13.11:
Obtain approximately the ratio of the nuclear radii of the gold isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev  and the silver isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev.
Ans: Nuclear radius of the gold isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev = RAu
Nuclear radius of the silver isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev = RAg
Mass number of gold, AAu = 197
Mass number of silver, AAg = 107
The ratio of the radii of the two nuclei is related with their mass numbers as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

Ques 13.12:
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) NCERT Solutions - Nuclei Class 12 Notes | EduRev and (b) NCERT Solutions - Nuclei Class 12 Notes | EduRev.

Given  NCERT Solutions - Nuclei Class 12 Notes | EduRev = 226.02540 u, 
NCERT Solutions - Nuclei Class 12 Notes | EduRev  = 222.01750 u,
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 220.01137 u,
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 216.00189 u.
Ans: (a) Alpha particle decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev 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.

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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 NCERT Solutions - Nuclei Class 12 Notes | EduRev

NCERT Solutions - Nuclei Class 12 Notes | EduRev

(b) Alpha particle decay of NCERT Solutions - Nuclei Class 12 Notes | EduRev  is shown by the following nuclear reaction.

NCERT Solutions - Nuclei Class 12 Notes | EduRev

It is given that:

Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 220.01137 u

Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 216.00189 u

Q-value = NCERT Solutions - Nuclei Class 12 Notes | EduRev

≈ 641 MeV

Kinetic energy of the α-particle NCERT Solutions - Nuclei Class 12 Notes | EduRev

= 6.29 MeV

Ques 13.13:
The radionuclide  11C decays according to
NCERT Solutions - Nuclei Class 12 Notes | EduRev
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

calculate Q and compare it with the maximum energy of the positron emitted
Ans: The given nuclear reaction is:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev = 11.011434 u
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the NCERT Solutions - Nuclei Class 12 Notes | EduRev  nucleus is given as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Where,
me = 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 me in the case of NCERT Solutions - Nuclei Class 12 Notes | EduRev and 5 me in the case of NCERT Solutions - Nuclei Class 12 Notes | EduRev.
Hence, equation (1) reduces to:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
∴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 Q is almost comparable to the maximum energy of the emitted positron.

Ques 13.14:
The nucleus NCERT Solutions - Nuclei Class 12 Notes | EduRev decays by NCERT Solutions - Nuclei Class 12 Notes | EduRev emission. Write down the NCERT Solutions - Nuclei Class 12 Notes | EduRev decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 22.994466 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 22.989770 u.
Ans: In  NCERT Solutions - Nuclei Class 12 Notes | EduRev emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
NCERT Solutions - Nuclei Class 12 Notes | EduRev emission of the nucleus NCERT Solutions - Nuclei Class 12 Notes | EduRev  is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 22.994466 u
Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
There are 10 electrons in  NCERT Solutions - Nuclei Class 12 Notes | EduRev and 11 electrons in NCERT Solutions - Nuclei Class 12 Notes | EduRev. 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 Me V/c2
∴ Q = 0.004696 x 931.5 = 4.374 Me V
The daughter nucleus is too heavy as compared to NCERT Solutions - Nuclei Class 12 Notes | EduRev and  NCERT Solutions - Nuclei Class 12 Notes | EduRev. 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.

Ques 13.15:
The Q value of a nuclear reaction A +  bC  + d is defined by Q = [ mA + mbmCmd]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.
(i) NCERT Solutions - Nuclei Class 12 Notes | EduRev
(ii) NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic masses are given to be
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Ans: (i) The given nuclear reaction is:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Atomic mass NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic mass NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic mass NCERT Solutions - Nuclei Class 12 Notes | EduRev
According to the question, the Q-value of the reaction can be written as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
The negative Q-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
The Q-value of this reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
The positive Q-value of the reaction shows that the reaction is exothermic.

Ques 13.16:
Suppose, we think of fission of a NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus into two equal fragments, NCERT Solutions - Nuclei Class 12 Notes | EduRev. Is the fission energetically possible? Argue by working out Q of the process. Given NCERT Solutions - Nuclei Class 12 Notes | EduRev and NCERT Solutions - Nuclei Class 12 Notes | EduRev.
Ans: The fission of NCERT Solutions - Nuclei Class 12 Notes | EduRev can be given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Atomic mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev = 55.93494 u
Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
The Q-value of this nuclear reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.


Ques 13.17:
The fission properties of NCERT Solutions - Nuclei Class 12 Notes | EduRev are very similar to those of NCERT Solutions - Nuclei Class 12 Notes | EduRevThe average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure NCERT Solutions - Nuclei Class 12 Notes | EduRev undergo fission?
Ans: Average energy released per fission of NCERT Solutions - Nuclei Class 12 Notes | EduRev, NCERT Solutions - Nuclei Class 12 Notes | EduRev
Amount of pure NCERT Solutions - Nuclei Class 12 Notes | EduRev, m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of NCERT Solutions - Nuclei Class 12 Notes | EduRev= 239 g
1 mole of NCERT Solutions - Nuclei Class 12 Notes | EduRev contains NA atoms.
m g of NCERT Solutions - Nuclei Class 12 Notes | EduRev containsNCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
∴Total energy released during the fission of 1 kg of NCERT Solutions - Nuclei Class 12 Notes | EduRev is calculated as:-
NCERT Solutions - Nuclei Class 12 Notes | EduRev

Hence, NCERT Solutions - Nuclei Class 12 Notes | EduRev  is released if all the atoms in 1 kg of pure NCERT Solutions - Nuclei Class 12 Notes | EduRev undergo fission.


Ques 13.18:
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much NCERT Solutions - Nuclei Class 12 Notes | EduRev did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of NCERT Solutions - Nuclei Class 12 Notes | EduRev and that this nuclide is consumed only by the fission process.
Ans: Half life of the fuel of the fission reactor,NCERT Solutions - Nuclei Class 12 Notes | EduRev  years
= 5 × 365 × 24 × 60 × 60 s
We know that in the fission of 1 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus, the energy released is equal to 200 MeV.
1 mole, i.e., 235 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev contains 6.023 × 1023 atoms.
∴1 g  NCERT Solutions - Nuclei Class 12 Notes | EduRev containsNCERT Solutions - Nuclei Class 12 Notes | EduRev
The total energy generated per gram of NCERT Solutions - Nuclei Class 12 Notes | EduRev is calculated as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
The reactor operates only 80% of the time.
Hence, the amount of NCERT Solutions - Nuclei Class 12 Notes | EduRev consumed in 5 years by the 1000 MW fission reactor is calculated as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

∴ Initial amount of NCERT Solutions - Nuclei Class 12 Notes | EduRev = 2 × 1538 = 3076 kg


Ques 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
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Ans: The given fusion reaction is:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

∴2.0 kg of deuterium contains NCERT Solutions - Nuclei Class 12 Notes | EduRev

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:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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:

NCERT Solutions - Nuclei Class 12 Notes | EduRev


Ques 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.)
Ans: When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1st deuteron Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm = 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:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

Where,
 NCERT Solutions - Nuclei Class 12 Notes | EduRev= Permittivity of free space
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

Ques 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).
Ans: We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0 = Constant.
A = Mass number of the nucleus
Nuclear matter density, NCERT Solutions - Nuclei Class 12 Notes | EduRev
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the nuclear matter density is independent of A. It is nearly constant.

Ques 13.22:
For the NCERT Solutions - Nuclei Class 12 Notes | EduRev  (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).

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Show that if NCERT Solutions - Nuclei Class 12 Notes | EduRev  emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.
Ans: Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev= Nuclear mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev= Nuclear mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev= Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev= Atomic mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev
me = Mass of an electron
c = Speed of light
Q-value of the electron capture reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Q-value of the positron capture reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
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 NCERT Solutions - Nuclei Class 12 Notes | EduRev 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.

Ques 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev (23.98504u), NCERT Solutions - Nuclei Class 12 Notes | EduRev (24.98584u) and NCERT Solutions - Nuclei Class 12 Notes | EduRev (25.98259u). The natural abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 78.99% by mass. Calculate the abundances of other two isotopes.
Ans: Average atomic mass of magnesium, m = 24.312 u
Mass of magnesium isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev, m1 = 23.98504 u
Mass of magnesium isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev, m2 = 24.98584 u
Mass of magnesium isotope NCERT Solutions - Nuclei Class 12 Notes | EduRev, m3 = 25.98259 u
Abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev, η1= 78.99%
Abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev, η2 = x%
Hence, abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev, η3 = 100 − x − 78.99% = (21.01 − x)%
We have the relation for the average atomic mass as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev

2431,2 = 1894.5783096 + 24.98584 x + 545.8942159 - 25.98259 x
0.99675* = 9.2725255
∴ x ≈ 9.3%
And 21.01 -* = 11.71%

Hence, the abundance of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 9.3% and that of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 11.71%.


Ques 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev and NCERT Solutions - Nuclei Class 12 Notes | EduRev from the following data:
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 39.962591 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev = 40.962278 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 25.986895 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev = 26.981541 u
Ans: For NCERT Solutions - Nuclei Class 12 Notes | EduRev
ForNCERT Solutions - Nuclei Class 12 Notes | EduRev
A neutron NCERT Solutions - Nuclei Class 12 Notes | EduRev is removed from a NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus. The corresponding nuclear reaction can be written as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Mass NCERT Solutions - Nuclei Class 12 Notes | EduRev = 39.962591 u
Mass NCERT Solutions - Nuclei Class 12 Notes | EduRev = 40.962278 u
Mass NCERT Solutions - Nuclei Class 12 Notes | EduRev = 1.008665 u
The mass defect of this reaction is given as:
Δm = NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev
∴Δm = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev

For NCERT Solutions - Nuclei Class 12 Notes | EduRev, the neutron removal reaction can be written as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is given that:
Mass  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 26.981541 u
Mass  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 25.986895 u
The mass defect of this reaction is given as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the energy required for neutron removal is calculated as:
NCERT Solutions - Nuclei Class 12 Notes | EduRev


Ques 13.25:
A source contains two phosphorous radio nuclides  NCERT Solutions - Nuclei Class 12 Notes | EduRev(T1/2 = 14.3d) and NCERT Solutions - Nuclei Class 12 Notes | EduRev (T1/2 = 25.3d). Initially, 10% of the decays come from NCERT Solutions - Nuclei Class 12 Notes | EduRev. How long one must wait until 90% do so?
Ans: Half life of NCERT Solutions - Nuclei Class 12 Notes | EduRev, T1/2 = 14.3 days
Half life of NCERT Solutions - Nuclei Class 12 Notes | EduRev, T’1/2 = 25.3 days
NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus decay is 10% of the total amount of decay.
The source has initially 10% of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus and 90% of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus.
Suppose after t days, the source has 10% of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus and 90% of  NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus.

Initially:
Number of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus = N
Number of NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus = 9 N
Finally:
Number of NCERT Solutions - Nuclei Class 12 Notes | EduRev
Number of NCERT Solutions - Nuclei Class 12 Notes | EduRev
For NCERT Solutions - Nuclei Class 12 Notes | EduRev nucleus, we can write the number ratio as:NCERT Solutions - Nuclei Class 12 Notes | EduRev

For NCERT Solutions - Nuclei Class 12 Notes | EduRev, we can write the number ratio as:

NCERT Solutions - Nuclei Class 12 Notes | EduRev

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

NCERT Solutions - Nuclei Class 12 Notes | EduRev

Hence, it will take about 208.5 days for 90% decay of  NCERT Solutions - Nuclei Class 12 Notes | EduRev.

Ques 13.26:
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

 NCERT Solutions - Nuclei Class 12 Notes | EduRev

 NCERT Solutions - Nuclei Class 12 Notes | EduRev

Calculate the Q-values for these decays and determine that both are energetically allowed.
Ans: Take a  NCERT Solutions - Nuclei Class 12 Notes | EduRev emission nuclear reaction:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
We know that:
Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRevm1 = 223.01850 u
Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRevm2 = 208.98107 u
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m3 = 14.00324 u
Hence, the Q-value of the reaction is given as:
Q = (m1m2m3) 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev emission nuclear reaction:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
We know that:
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRevm1 = 223.01850
Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRevm2 = 219.00948
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m3 = 4.00260
Q-value of this nuclear reaction is given as:
Q = (m1m2m3) c2
= (223.01850 − 219.00948 − 4.00260) C2
= (0.00642 c2) u
= 0.00642 × 931.5 = 5.98 MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

Ques 13.27:
Consider the fission of NCERT Solutions - Nuclei Class 12 Notes | EduRev 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 NCERT Solutions - Nuclei Class 12 Notes | EduRev  and NCERT Solutions - Nuclei Class 12 Notes | EduRev. Calculate Q for this fission process. The relevant atomic and particle masses are
mNCERT Solutions - Nuclei Class 12 Notes | EduRev  =238.05079 u
mNCERT Solutions - Nuclei Class 12 Notes | EduRev  =139.90543 u
mNCERT Solutions - Nuclei Class 12 Notes | EduRev  = 98.90594 u

Ans: In the fission of NCERT Solutions - Nuclei Class 12 Notes | EduRev, 10 β particles decay from the parent nucleus. The nuclear reaction can be written as:

 NCERT Solutions - Nuclei Class 12 Notes | EduRev

It is given that:
Mass of a nucleus NCERT Solutions - Nuclei Class 12 Notes | EduRevm1 = 238.05079 u
Mass of a nucleus  NCERT Solutions - Nuclei Class 12 Notes | EduRevm2 = 139.90543 u
Mass of a nucleus NCERT Solutions - Nuclei Class 12 Notes | EduRev, m3 = 98.90594 u
Mass of a neutron NCERT Solutions - Nuclei Class 12 Notes | EduRevm4 = 1.008665 u
Q-value of the above equation,
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Where,
m’ = Represents the corresponding atomic masses of the nuclei
NCERT Solutions - Nuclei Class 12 Notes | EduRev= m1 − 92me
NCERT Solutions - Nuclei Class 12 Notes | EduRev= m2 − 58me
NCERT Solutions - Nuclei Class 12 Notes | EduRev= m3 − 44me
NCERT Solutions - Nuclei Class 12 Notes | EduRev= m4

NCERT Solutions - Nuclei Class 12 Notes | EduRev

But 1 u = 931.5 MeV / c2
∴ O = 0.247995 x 931.5 = 23 1.007 MeV

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

Ques 13.28:
Consider the D−T reaction (deuterium−tritium fusion)
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
(a) Calculate the energy released in MeV in this reaction from the data:
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 2.014102 u
NCERT Solutions - Nuclei Class 12 Notes | EduRev= 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 = Boltzmann’s constant, T = absolute temperature.)
Ans: (a) Take the D-T nuclear reaction: NCERT Solutions - Nuclei Class 12 Notes | EduRev 
It is given that:
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m1= 2.014102 u
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m2 = 3.016049 u
Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRevm3 = 4.002603 u
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev, m4 = 1.008665 u
Q-value of the given D-T reaction is:
Q = [m1  + m2m3 − 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 × 10−15 m
Distance between the two nuclei at the moment when 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 between the two nuclei is given as:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Where,
0 = Permittivity of free space
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, 5.76 × 10−14 J or NCERT Solutions - Nuclei Class 12 Notes | EduRev of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.
However, it is given that:
KENCERT Solutions - Nuclei Class 12 Notes | EduRev
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1
T = Temperature required for triggering the reaction
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.

Ques 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

 NCERT Solutions - Nuclei Class 12 Notes | EduRev

Ans: It can be observed from the given γ-decay diagram that γ1 decays 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
1= 1.088 × 1.6 × 10−19 × 106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiation radiated by γ1-decay
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
It can be observed from the given γ-decay diagram that γ2 decays 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
2= 0.412 × 1.6 × 10−19 × 106 J
Where,
ν2 = Frequency of radiation radiated by γ2-decay
 NCERT Solutions - Nuclei Class 12 Notes | EduRev

It can be observed from the given γ-decay diagram that γ3 decays 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
3= 0.676 × 10−19 × 106 J
Where,
ν3 = Frequency of radiation radiated by γ3-decay
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Mass of NCERT Solutions - Nuclei Class 12 Notes | EduRev = 197.968233 u
Mass of  NCERT Solutions - Nuclei Class 12 Notes | EduRev= 197.966760 u
1 u = 931.5 MeV/c2
Energy of the highest level is given as:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
∴Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
∴Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV

Ques 13.30:
Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.
Ans:  (a) Amount of hydrogen, m = 1 kg = 1000 g
1 mole, i.e., 1 g of hydrogen (NCERT Solutions - Nuclei Class 12 Notes | EduRev ) contains 6.023 × 1023 atoms.
∴1000 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev  contains 6.023 × 1023 × 1000 atoms.
Within the sun, four  NCERT Solutions - Nuclei Class 12 Notes | EduRev nuclei combine and form one NCERT Solutions - Nuclei Class 12 Notes | EduRev  nucleus. In this process 26 MeV of energy is released.
Hence, the energy released from the fusion of 1 kg NCERT Solutions - Nuclei Class 12 Notes | EduRev is:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
(b) Amount of  NCERT Solutions - Nuclei Class 12 Notes | EduRev = 1 kg = 1000 g
1 mole, i.e., 235 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev  contains 6.023 × 1023 atoms.
∴1000 g of NCERT Solutions - Nuclei Class 12 Notes | EduRev contains NCERT Solutions - Nuclei Class 12 Notes | EduRev
It is known that the amount of energy released in the fission of one atom of NCERT Solutions - Nuclei Class 12 Notes | EduRev is 200 MeV.
Hence, energy released from the fission of 1 kg of NCERT Solutions - Nuclei Class 12 Notes | EduRev  is:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
NCERT Solutions - Nuclei Class 12 Notes | EduRev

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.

Ques 13.31: 
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an 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.
Ans: Amount of electric power to be generated, P = 2 × 105 MW
10% of this amount has to be obtained from nuclear power plants.
∴ Amount of nuclear power, NCERT Solutions - Nuclei Class 12 Notes | EduRev
= 2 × 104 MW
= 2 × 104 × 106 J/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:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Number of atoms required for fission per year:
 NCERT Solutions - Nuclei Class 12 Notes | EduRev

1 mole, i.e., 235 g of U235 contains 6.023 × 1023 atoms.
∴ Mass of 6.023 × 1023 atoms of U235 = 235 g = 235 × 10−3 kg
∴ Mass of 78840 × 1024 atoms of U235
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
 NCERT Solutions - Nuclei Class 12 Notes | EduRev
Hence, the mass of uranium needed per year is 3.076 × 104 kg.

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