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NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

NCERT QUESTION
(Nuclei)
Q13.1: Obtain the binding energy (in MeV) of a nitrogen nucleus NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, given NCERT Solutions Class 12 Physics Chapter 13 - Nuclei =14.00307 u
Ans: Atomic mass of nitrogen NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m = 14.00307 u
A nucleus of nitrogen NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  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 Class 12 Physics Chapter 13 - Nuclei
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

Q13.2: Obtain the binding energy of the nuclei NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei in units of MeV from the following data:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 55.934939 u , NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 208.980388 u
Ans: Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m1 = 55.934939 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei
= 492.26 MeV
Average binding energy per nucleon NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m2 = 208.980388 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei
= 1640.26 MeV
Average binding energy per nucleon = NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

Q13.3: 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 Class 12 Physics Chapter 13 - Nuclei atoms (of mass 62.92960 u).
Ans: Mass of a copper coin, m’ = 3 g
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei atom, m = 62.92960 u
The total number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei atoms in the coinNCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nucleinucleus 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 Class 12 Physics Chapter 13 - Nuclei
= 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.

Q13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  and the silver isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.
Ans: Nuclear radius of the gold isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = RAu
Nuclear radius of the silver isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 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 Class 12 Physics Chapter 13 - Nuclei
Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Q13.5: 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 Class 12 Physics Chapter 13 - Nuclei
(ii) NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic masses are given to be
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Ans: (i) The given nuclear reaction is:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Atomic mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
According to the question, the Q-value of the reaction can be written as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The negative Q-value of the reaction shows that the reaction is endothermic.
(ii) The given nuclear reaction is:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The Q-value of this reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The positive Q-value of the reaction shows that the reaction is exothermic.

Q13.6: Suppose, we think of fission of a NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus into two equal fragments, NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. Is the fission energetically possible? Argue by working out Q of the process. Given NCERT Solutions Class 12 Physics Chapter 13 - Nuclei and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.
Ans: The fission of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei can be given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 55.93494 u
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The Q-value of this nuclear reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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.

Q13.7: The fission properties of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei are very similar to those of NCERT Solutions Class 12 Physics Chapter 13 - NucleiThe 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 Class 12 Physics Chapter 13 - Nuclei undergo fission?
Ans: Average energy released per fission of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Amount of pure NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 239 g
1 mole of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei contains NA atoms.
m g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei containsNCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
∴Total energy released during the fission of 1 kg of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is calculated as:-
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  is released if all the atoms in 1 kg of pure NCERT Solutions Class 12 Physics Chapter 13 - Nuclei undergo fission.

Q13.8: 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 Class 12 Physics Chapter 13 - Nuclei
Ans: The given fusion reaction is:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei

Q13.9: 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 Class 12 Physics Chapter 13 - Nuclei
Where,
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= Permittivity of free space
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

Q13.10: 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 Class 12 Physics Chapter 13 - Nuclei
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the nuclear matter density is independent of A. It is nearly constant.


Old NCERT Solutions

Q1: For the NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  (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 Class 12 Physics Chapter 13 - Nuclei
Show that if NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  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 Class 12 Physics Chapter 13 - Nuclei
Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= Nuclear mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= Nuclear mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
me = Mass of an electron
c = Speed of light
Q-value of the electron capture reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Q-value of the positron capture reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei 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.

Q2: 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 Class 12 Physics Chapter 13 - Nuclei (23.98504u), NCERT Solutions Class 12 Physics Chapter 13 - Nuclei (24.98584u) and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei (25.98259u). The natural abundance of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei, m1 = 23.98504 u
Mass of magnesium isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m2 = 24.98584 u
Mass of magnesium isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m3 = 25.98259 u
Abundance of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η1= 78.99%
Abundance of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η2 = x%
Hence, abundance of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η3 = 100 − x − 78.99% = (21.01 − x)%
We have the relation for the average atomic mass as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei is 9.3% and that of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 11.71%.

Q3: 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 Class 12 Physics Chapter 13 - Nuclei and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei from the following data:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 39.962591 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 40.962278 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 25.986895 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 26.981541 u
Ans: For NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
ForNCERT Solutions Class 12 Physics Chapter 13 - Nuclei
A neutron NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is removed from a NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus. The corresponding nuclear reaction can be written as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 39.962591 u
Mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 40.962278 u
Mass NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 1.008665 u
The mass defect of this reaction is given as:
Δm = NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
∴Δm = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
For NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, the neutron removal reaction can be written as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Mass  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 26.981541 u
Mass  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 25.986895 u
The mass defect of this reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the energy required for neutron removal is calculated as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

Q4: A source contains two phosphorous radio nuclides  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei(T1/2 = 14.3d) and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei (T1/2 = 25.3d). Initially, 10% of the decays come from NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. How long one must wait until 90% do so?
Ans: Half life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, T1/2 = 14.3 days
Half life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, T’1/2 = 25.3 days
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus decay is 10% of the total amount of decay.
The source has initially 10% of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus and 90% of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus.
Suppose after t days, the source has 10% of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus and 90% of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus.
Initially:
Number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus = N
Number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus = 9 N
Finally:
Number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Number of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
For NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus, we can write the number ratio as:NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
For NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, we can write the number ratio as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
On dividing equation (1) by equation (2), we get:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, it will take about 208.5 days for 90% decay of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.

Q5: Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Calculate the Q-values for these decays and determine that both are energetically allowed.
Ans: Take a  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei emission nuclear reaction:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
We know that:
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nucleim1 = 223.01850 u
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nucleim2 = 208.98107 u
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, 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 Class 12 Physics Chapter 13 - Nuclei emission nuclear reaction:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
We know that:
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nucleim1 = 223.01850
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nucleim2 = 219.00948
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, 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.

Q6: Consider the fission of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei  and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. Calculate Q for this fission process. The relevant atomic and particle masses are
mNCERT Solutions Class 12 Physics Chapter 13 - Nuclei  =238.05079 u
mNCERT Solutions Class 12 Physics Chapter 13 - Nuclei  =139.90543 u
mNCERT Solutions Class 12 Physics Chapter 13 - Nuclei  = 98.90594 u
Ans: In the fission of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, 10 β particles decay from the parent nucleus. The nuclear reaction can be written as:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Mass of a nucleus NCERT Solutions Class 12 Physics Chapter 13 - Nucleim1 = 238.05079 u
Mass of a nucleus  NCERT Solutions Class 12 Physics Chapter 13 - Nucleim2 = 139.90543 u
Mass of a nucleus NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m3 = 98.90594 u
Mass of a neutron NCERT Solutions Class 12 Physics Chapter 13 - Nucleim4 = 1.008665 u
Q-value of the above equation,
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Where,
m’ = Represents the corresponding atomic masses of the nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= m1 − 92me
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= m2 − 58me
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= m3 − 44me
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= m4
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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.

Q7: Consider the D−T reaction (deuterium−tritium fusion)
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(a) Calculate the energy released in MeV in this reaction from the data:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 2.014102 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 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 Class 12 Physics Chapter 13 - Nuclei 
It is given that:
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m1= 2.014102 u
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m2 = 3.016049 u
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nucleim3 = 4.002603 u
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, 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 Class 12 Physics Chapter 13 - Nuclei
Where,
0 = Permittivity of free space
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, 5.76 × 10−14 J or NCERT Solutions Class 12 Physics Chapter 13 - Nuclei of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.
However, it is given that:
KENCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1
T = Temperature required for triggering the reaction
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.

Q8: 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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 197.968233 u
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 197.966760 u
1 u = 931.5 MeV/c2
Energy of the highest level is given as:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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

Q9: 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 Class 12 Physics Chapter 13 - Nuclei ) contains 6.023 × 1023 atoms.
∴1000 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  contains 6.023 × 1023 × 1000 atoms.
Within the sun, four  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nuclei combine and form one NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  nucleus. In this process 26 MeV of energy is released.
Hence, the energy released from the fusion of 1 kg NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(b) Amount of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 1 kg = 1000 g
1 mole, i.e., 235 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  contains 6.023 × 1023 atoms.
∴1000 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei contains NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is known that the amount of energy released in the fission of one atom of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 200 MeV.
Hence, energy released from the fission of 1 kg of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  is:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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.

Q10: 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 Class 12 Physics Chapter 13 - Nuclei
= 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 Class 12 Physics Chapter 13 - Nuclei
Number of atoms required for fission per year:
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
 NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the mass of uranium needed per year is 3.076 × 104 kg.

Q11: (a) Two stable isotopes of lithium NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. 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 Class 12 Physics Chapter 13 - Nuclei  and NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.
Ans: (a) Mass of lithium isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei , m1 = 6.01512 u
Mass of lithium isotope  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei,  m2 = 7.01600 u
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η1= 7.5%
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η2= 92.5%
The atomic mass of lithium atom is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(b) Mass of boron isotope  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m1 = 10.01294 u
Mass of boron isotope  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m2 = 11.00931 u
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η1 = x%
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η2= (100 − x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
And 100 − x = 80.11%
Hence, the abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 19.89% and that of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 80.11%.

Q12: The three stable isotopes of neon:  NCERT Solutions Class 12 Physics Chapter 13 - Nucleiand NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei, m1= 19.99 u
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η1 = 90.51%
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m2 = 20.99 u
Abundance of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei , η2 = 0.27%
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, m3 = 21.99 u
Abundance of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, η3 = 9.22%
The average atomic mass of neon is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
  = 20.1771 u

Q13: Write nuclear reaction equations for
(i) α-decay of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(ii) α-decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(iii) β-decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(iv) β-decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(v) β -decay of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(vi) β -decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(vii) Electron capture of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Ans: α is a nucleus of helium NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

Q14: 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 Class 12 Physics Chapter 13 - Nuclei
Where,
λ = Decay constant
t = Time
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
Since, λ = 0.693/T
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the isotope will take about 6.645 T years to reduce to 1% of its original value.

Q15: 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 Class 12 Physics Chapter 13 - Nuclei  present with the stable carbon isotope NCERT Solutions Class 12 Physics Chapter 13 - Nuclei . 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 Class 12 Physics Chapter 13 - Nuclei, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei,  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 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 Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.

Q16: Obtain the amount of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei necessary to provide a radioactive source of 8.0 mCi strength. The half-life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 5.3 years.
Ans: The strength of the radioactive source is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Where,
N = Required number of atoms
Half-life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei, NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  = 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 Class 12 Physics Chapter 13 - Nuclei
Where, λ NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
For NCERT Solutions Class 12 Physics Chapter 13 - Nuclei:
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
∴Mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei atoms NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the amount of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  necessary for the purpose is 7.106 × 10−6 g.

Q17: The half-life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is 28 years. What is the disintegration rate of 15 mg of this isotope?
Ans: Half life of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei , NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 28 years
= 28 × 365 × 24 × 60 × 60
= 8.83 × 108 s
Mass of the isotope, m = 15 mg
90 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei atom contains 6.023 × 1023 (Avogadro’s number) atoms.
Therefore, 15 mg of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei contains:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Rate of disintegration, NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Where,
λ = Decay constant NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 1010 atoms/s.

Q18: Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) NCERT Solutions Class 12 Physics Chapter 13 - Nuclei and (b) NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.
Given  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 226.02540 u, 
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  = 222.01750 u,
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 220.01137 u,
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 216.00189 u.
Ans: (a) Alpha particle decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
(b) Alpha particle decay of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  is shown by the following nuclear reaction.
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 220.01137 u
Mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 216.00189 u
Q-value = NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
≈ 641 MeV
Kinetic energy of the α-particle NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
= 6.29 MeV

Q19: The radionuclide  11C decays according to
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
calculate Q and compare it with the maximum energy of the positron emitted
Ans: The given nuclear reaction is:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 11.011434 u
Atomic mass of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei  nucleus is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
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 Class 12 Physics Chapter 13 - Nuclei and 5 me in the case of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei.
Hence, equation (1) reduces to:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
∴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.

Q20: The nucleus NCERT Solutions Class 12 Physics Chapter 13 - Nuclei decays by NCERT Solutions Class 12 Physics Chapter 13 - Nuclei emission. Write down the NCERT Solutions Class 12 Physics Chapter 13 - Nuclei decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 22.994466 u
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 22.989770 u.
Ans: In  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei emission of the nucleus NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
It is given that:
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 22.994466 u
Atomic mass of  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei= 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
There are 10 electrons in  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei and 11 electrons in NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. 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 Class 12 Physics Chapter 13 - Nuclei and  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei. 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.

Q21: A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much NCERT Solutions Class 12 Physics Chapter 13 - Nuclei 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 Class 12 Physics Chapter 13 - Nuclei and that this nuclide is consumed only by the fission process.
Ans: Half life of the fuel of the fission reactor,NCERT Solutions Class 12 Physics Chapter 13 - Nuclei  years
= 5 × 365 × 24 × 60 × 60 s
We know that in the fission of 1 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei nucleus, the energy released is equal to 200 MeV.
1 mole, i.e., 235 g of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei contains 6.023 × 1023 atoms.
∴1 g  NCERT Solutions Class 12 Physics Chapter 13 - Nuclei containsNCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The total energy generated per gram of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is calculated as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
The reactor operates only 80% of the time.
Hence, the amount of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei consumed in 5 years by the 1000 MW fission reactor is calculated as:
NCERT Solutions Class 12 Physics Chapter 13 - Nuclei
∴ Initial amount of NCERT Solutions Class 12 Physics Chapter 13 - Nuclei = 2 × 1538 = 3076 kg.

The document NCERT Solutions Class 12 Physics Chapter 13 - Nuclei is a part of the NEET Course Physics Class 12.
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FAQs on NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

1. What is a nucleus in the context of atomic structure?
Ans. The nucleus is the central part of an atom that contains protons and neutrons, which are positively and neutrally charged subatomic particles, respectively.
2. How does the nucleus contribute to the mass of an atom?
Ans. The nucleus is where the majority of an atom's mass is located due to the presence of protons and neutrons, which have much greater mass than the electrons that orbit the nucleus.
3. What is nuclear fusion and how does it relate to the nucleus?
Ans. Nuclear fusion is a process where multiple atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. This process is related to the nucleus as it involves changes within the atomic nuclei.
4. How does the structure of the nucleus affect an atom's stability?
Ans. The number of protons and neutrons in the nucleus determines the stability of an atom. If the nucleus is too large or too small, it can lead to instability and radioactive decay.
5. Can the nucleus of an atom be altered through nuclear reactions?
Ans. Yes, the nucleus of an atom can be altered through nuclear reactions such as fission and fusion, which involve splitting or combining atomic nuclei to release energy.
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