NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei

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CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26

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Have you ever asked yourself how the stars manage to generate such a great amount of energy, or how nuclear power stations manage to produce electricity? The solution is in the minute nucleus of atoms, which, in spite of being almost a thousand times smaller than the atom itself, contains nearly all of its weight. Class 12 Physics Chapter 13 Nuclei presents the students with the fascinating field of nuclear physics, which includes such topics as atomic masses, nuclear binding energy, nuclear forces, radioactivity, and nuclear energy, which is fundamental to the field in both theory and practice.

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  1. NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Download PDF
  2. NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Exercise Questions
  3. NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Additional Questions
  4. NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Higher Order Thinking Skills (HOTS) Questions
  5. Nuclei NCERT Solutions: Topics
  6. Class 12 Physics Chapter 13 - Nuclei: Important Formulas
  7. Approach to Solve Questions of Class 12 Physics Chapter 13 - Nuclei
  8. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  9. NCERT Solutions for Class 12 Physics Chapter-wise
NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei
Nuclie

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei are specifically developed by the experts of the subject to explain all the exercises given in the textbook, the other practice questions, and the HOTS problems step by step. These NCERT solutions have made complex nuclear concepts simple to understand, and at the same time, derivations and solving of numerical problems are easy to understand. These NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei are available in free PDF and are highly useful in the preparation of the board exams of Class 12 of the CBSE, and a powerful resource for JEE and NEET candidates. Through these solutions, students will be able to improve their fundamentals, gain accuracy, and gain confidence in the exam.

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Download PDF

The Class 12 Physics Chapter 13 - Nuclei Solutions presented the answers to all the questions with step-by-step explanations. These solutions will make the students grasp tricky topics without difficulty and can be used both in the CBSE board exams as well as JEE and NEET exams. The PDF version can also be downloaded so that you can have an easy offline reference and a quick revision anytime.

Download PDF

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Exercise Questions

Class 12 Physics Chapter 12 - Nuclei question answers (Exercise questions) offer step-by-step answers to all textbook questions, covering topics like nuclear structure, binding energy, and radioactivity. These solutions help students strengthen concepts, practice effectively, and prepare for board exams as well as competitive exams like JEE and NEET.

13.1 Obtain the binding energy( in MeV ) of a nitrogen nucleus (714N) , given m (714N)=14.00307u

Answer:

mn=1.00866u mp=1.00727u
Atomic mass of Nitrogen m=14.00307u
Mass defect Δm=7×mn+7×mpm

Δm=7×1.00866+7×1.0072714.00307

Δm=0.10844
Now 1u is equivalent to 931.5 MeV

Eb=0.10844×931.5Eb=101.01186MeV
Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV

13.2 (i) Obtain the binding energy of the nuclei 2656Fe and 83209Bi in units of MeV from the following data:

m(2656Fe)=55.934939u

Answer:

mH=1.007825umn=1.008665u


Atomic mass of 2656Fe is m=55.934939u
Mass defect

Δm=(5626)×mH+26×mpmΔm=30×1.008665+26×1.00782555.934939Δm=0.528461


Now 1 u is equivalent to 931.5 MeV

Eb=0.528461×931.5Eb=492.2614215MeV


Therefore binding energy of a 2656Fe nucleus is 492.2614215 MeV.
Average binding energy

=492.2656MeV=8.79MeV

13.2 (ii) Obtain the binding energy of the nuclei 2656Fe and 83209Bi in units of MeV from the following data:

m(83209Bi)=208.980388u

Answer:

mH=1.007825u


mn=1.008665u


Atomic mass of 83209Bi is m=208.980388u
Mass defect

Δm=(20983)+83×mHmΔm=126×1.008665+83×1.007825208.980388Δm=1.760877u


Now 1u is equivalent to 931.5 MeV

Eb=1.760877×931.5Eb=1640.2569255MeV


Therefore binding energy of a 83209Bi nucleus is 1640.2569255 MeV .
Average binding energy =1640.25208.98=7.84MeV

13.3 A given coin has a mass of 3.0g. 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 2963Cu atoms (of mass 62.92960u ).

Answer:

Mass of the coin is w=3g
Total number of Cu atoms in the coin is n

n=w×NA Atomic Mass n=3×6.023×102362.92960n=2.871×1022 mH=1.007825u mn=1.008665u


Atomic mass of 2963Cu is m=62.92960u
Mass defect Δm=(6329)×mn+29×mHm
Δm=34×1.008665+29×1.00782562.92960
Δm=0.591935u
Now 1u is equivalent to 931.5 MeV

Eb=0.591935×931.5

Eb=551.38745MeV
Therefore binding energy of a 2963Cu nucleus is 551.38745 MeV.
The nuclear energy that would be required to separate all the neutrons and protons from each other is

n×Eb=2.871×1022×551.38745=1.5832×1025MeV=1.5832×1025×1.6×1019×106 J=2.5331×109 kJ

13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope 79197Au and the silver isotope 47107Ag

Answer:

The nuclear radii are directly proportional to the cube root of the mass number.

The ratio of the radii of the given isotopes is

(197107)1/3=1.23

13.5 (i) 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.

11H+13H12H+12H the following

Atomic masses are given to be

m(12H)=2.014102u

m(13H)=3.0016049u

m(612H)=12.000000u

m(1020Ne)=19.992439u

Answer:

Δm=m(11H)+m(13H)2m(12H)
Δm=1.007825+3.00160492×2.014102
Δm=0.00433

The above negative value of mass defect implies there will be a negative Q value, and therefore the reaction is endothermic

13.5 (ii) 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.

612C+612C1020Ne+24He

Atomic masses are given to be

m(12H)=2.014102u

m(13H)=3.0016049u

m(612H)=12.000000u

m(1020Ne)=19.992439u

Answer:

Δm=2m(612C)m(1020Ne)m(24He)
Δm=2×12.0000019.9924394.002603
Δm=0.004958

The above positive value of mass defect implies the Q value would be positive and therefore the reaction is exothermic.

13.6 Suppose, we think of fission of a 2656Fe nucleus into two equal fragments, 1328Al . Is the fission energetically possible? Argue by working out Q of the process. Given m(2656Fe)=55.93494u and m(1328Al)=27.98191u

Answer:

The reaction will be 2656Fe1328Al+1328Al

The mass defect of the reaction will be

Δm=m(2656Fe)2m(1328Al)
Δm=55.934942×27.98191
Δm=0.02888u

Since the mass defect is negative, the Q value will also be negative, and therefore, the fission is not energetically possible.

13.7 The fission properties of 94239Pu are very similar to those of 92235U . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 94239Pu undergo fission?

Answer:

Number of atoms present in 1 kg(w) of 94239Pu=n

n=w×NA mass number of Pu n=1000×6.023×1023239n=2.52×1024


Energy per fission (E)=180MeV
Total Energy released if all the atoms in 1 kg94239Pu undergo fission =E×n

=180×2.52×1024=4.536×1026MeV

13.8 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

12H+12H23He+n+3.27MeV

Answer:

The energy liberated on the fusion of two atoms of deuterium is 3.27 MeV

Number of fusion reactions in 2 kg of deuterium = N A × 500

The energy liberated by the fusion of 2.0 kg of deuterium atoms E

=3.27×106×1.6×1019×6.023×1023×500=1.576×1014 J

Power of lamp (P)= 100 W

The time for which the lamp would glow using E amount of energy is

T=EP=1.576×1014100×3600×24×365

=4.99 × 104 years

13.9 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.)

Answer:

For a head-on collision of two deuterons, the closest distances between their centres will be d=2×r

d=2×2.0d=4.0fmd=4×1015 m

charge on each deuteron = charge of one proton =q=1.6×1019C
The maximum electrostatic potential energy of the system during the head-on collision will be E

=q24πϵ0d=9×109×(1.6×1019)24×1015J=9×109×(1.6×1019)24×1015×1.6×1019eV=360keV


The above basically means that bringing two deuterons from infinity to each other would require 360 keV of work to be done or would require 360 keV of energy to be spent.

13.10 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).

Answer:

The mass of an element with mass number A will be about A u. The density of its nucleus, therefore, would be

d=mvd=A4π3R3d=A4π3(R0A1/3)3d=34πR03


As we can see, the above density comes out to be independent of mass number A, and R0 is constant, so matter density is nearly constant

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Additional Questions

Nuclei class 12 question answers (Additional Questions) are designed to enhance conceptual clarity and problem-solving skills beyond the NCERT textbook. These questions cover nuclear forces, energy, and decay processes, making them highly useful for board exams and entrance tests like JEE/NEET.

1.(a) Two stable isotopes of lithium 36Li and 37Li have respective abundances of 7.5o/o and 92.5o/o . These isotopes have masses 6.01512u and 7.01600u , respectively. Find the atomic mass of lithium.

Answer:

Mass of the two stable isotopes and their respective abundances are 6.01512u and 7.01600u and 7.5o/o and 92.5o/o .

m=6.01512×7.5+7.01600×92.5100

m=6.940934 u

1.(b) Boron has two stable isotopes, 510B and 511B . Their respective masses are 10.01294u and 11.00931u , and the atomic mass of boron is 10.811 u. Find the abundances of 510B and 511B .

Answer:

The atomic mass of boron is 10.811 u

Mass of the two stable isotopes are 10.01294u and 11.00931u respectively
Let the two isotopes have abundances x% and (100x)%

10.811=10.01294×x+11.00931×(100x)100x=19.89100x=80.11


Therefore the abundance of 510 B is 19.89% and that of 511 B is 80.11%

2. The three stable isotopes of neon: 1020Ne, 1021Ne and 1022Ne have respective abundances of 90.51o/o , 0.27o/o and 9.22o/o . The atomic masses of the three isotopes are 19.99u,20.99uand21.99u, respectively. Obtain the average atomic mass of neon.

Answer:

The atomic masses of the three isotopes are 19.99u(m1),20.99u(m2) and 21.99u(m3)
Their respective abundances are 90.51%(p1),0.27%(p2) and 9.22%(p3)

m=19.99×90.51+20.99×0.27+21.99×9.22100m=20.1771u

The average atomic mass of neon is 20.1771 u .

3. (i) Write nuclear reaction equations for

αdecayof88226Ra

Answer:

The nuclear reaction equations for the given alpha decay

88226Ra86222Rn+24He

3. (ii) Write nuclear reaction equations for

αdecayof94242Pu

Answer:

The nuclear reaction equations for the given alpha decay is

94242Pu92238U+24He

3. (iii) Write nuclear reaction equations for

βdecayof1532P

Answer:

The nuclear reaction equations for the given beta minus decay is

1532P1632S+e+ν¯

3.(iv) Write nuclear reaction equations for

βdecayof83210Bi

Answer:

The nuclear reaction equation for the given beta minus decay is

83210Bi84210Po+e+ν¯

3.(v) Write nuclear reaction equations for

β+decayof611C

Answer:

The nuclear reaction for the given beta plus decay will be

611C511P+e++ν

3.(vi) Write nuclear reaction equations for

β+decayof4397Tc

Answer:

nuclear reaction equations for

β+decayof4397Tc is

4397Tc4297Mo+e++ν

3.(vii) Write nuclear reaction equations for

Electron capture of 54120Xe

Answer:

The nuclear reaction for electron capture of 54120Xe is

54120Xe+e53120I+ν

4. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, and b) 1% of its original value?

Answer:

(a) The activity is proportional to the number of radioactive isotopes present

The number of half years in which the number of radioactive isotopes reduces to x% of its original value is n.

n=log2(100x)

In this case

n=log2(1003.125)=log232=5

It will take 5T years to reach 3.125% of the original activity.

(b) In this case

n=log2(1001)=log2100=6.64

It will take 6.64T years to reach 1% of the original activity.

5. 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 614C present with the stable carbon isotope 612C . 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 614C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 614C 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.

Answer:

Since we know that activity is proportional to the number of radioactive isotopes present in the sample.

RR0=NN0=915=0.6


Also

N=N0eλtt=1λlnNN0t=1λln0.6t=0.51λ

but λ=0.693T1/2
Therefore

t=0.51×T1/20.693t=0.735T1/2t4217


The age of the Indus-Valley civilisation, calculated using the given specimen, is approximately 4217 years.

6. Obtain the amount of 2760Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 2760Co is 5.3 years.

Answer:

Required activity =8.0mCi
1Ci=3.7×1010 decay s s1
8.0mCi=8×103×3.7×1010=2.96×108 decay s 1
T1/2=5.3 years

λ=0.693T1/2


λ=0.6935.3×365×24×3600


λ=4.14×109s1


dN dt=Nλ


N=dN dt×1λ


N=(2.96×108)×14.14×109


N=7.15×1016 atoms


Mass of those many atoms of Cu will be

w=7.15×1016×606.023×1023w=7.12×106g

7.12×106 g of 2760Co is necessary to provide a radioactive source of 8.0 mCi strength.

7. The half-life of 3890Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer:

T1/2=28 years

λ=0.69328×365×24×3600λ=7.85×1010 decay s1


The number of atoms in 15 mg of 3890Sr is

N=15×103×6.023×102390N=1.0038×1020


The disintegration rate will be

dN dt=Nλ=1.0038×1020×7.85×1010=7.88×1010 s1


The disintegration rate is therefore 7.88×1010 decay s1.

8.(a) Find the Q-value and the kinetic energy of the emitted α -particle in the α -decay of


88226Ra

Given m(88226Ra)=226.02540u, m(86222Rn)=222.01750u,

m(86222Rn)=220.01137u, m(84216Po)=216.00189u,

Answer:

Mass defect is Δm

Δm=m(88226Ra)m(86222Rn)m(24He)Δm=226.02540222.01754.002603Δ m=0.005297u1u=931.5MeV/c2 Q-value =Δm×931.5=4.934515MeV


By using Linear Momentum Conservation and Energy Conservation
Kinetic energy of alpha particle =

mass of nucleus after decay mass of nucleus before decay ×Qvalue =222.01750226.0254×4.934515=4.847MeV

8.(b) Find the Q-value and the kinetic energy of the emitted α -particle in the α -decay of


86220Rn

Given m(88226Ra)=226.02540u, m(86222Rn)=222.01750u,

m(86222Rn)=220.01137u, m(84216Po)=216.00189u,

Answer:

Mass defect is Δm

Δm=m(88226Ra)m(86222Rn)m(24He)Δm=226.02540222.01754.002603Δ m=0.005297u1u=931.5MeV/c2 Q-value =Δm×931.5=4.934515MeV


By using Linear Momentum Conservation and Energy Conservation
Kinetic energy of alpha particle =

mass of nucleus after decay mass of nucleus before decay ×Q value =222.01750226.0254×4.934515=4.847MeV

9. The radionuclide 11C decays according to

611CB+e++v:T1/2=20.3min
The maximum energy of the emitted positron is 0.960MeV. .

Given the mass values:

m(611C)=11.011434u and m(611B)=11.009305u
calculate Q and compare it with the maximum energy of the positron emitted.

Answer:

If we use atomic masses

Δm=m(611C)m(511B)2me
Δm=11.01143411.0093052×0.000548
Δm=0.001033u

Q-value= 0.001033 × 931.5=0.9622 MeV, which is comparable with the maximum energy of the emitted positron.

10. The nucleus 1023Ne decays by β emission. Write down the β -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

(i)m(1023Ne)=22.994466u

(ii)m(1123Na)=22.089770u

Answer:

The β decay equation is

1023Ne1123Na+e+ν¯+Q

Δm=m(1023Ne)1123Name
Δm=22.99446622.989770
Δm=0.004696u

(we did not subtract the mass of the electron, as it is cancelled because of the presence of one more electron in the sodium atom)

Q=0.004696 × 931.5

Q=4.3743 eV

The emitted nucleus is way heavier than the β particle and the energy of the antineutrino is also negligible therefore, the maximum energy of the emitted electron is equal to the Q value.

11. A 1000MW fission reactor consumes half of its fuel in 5.00y . How much 92235U did it contain initially? Assume that the reactor operates 80o/0 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:

The amount of energy liberated on fission of 1 92235U atom is 200 MeV.

The amount of energy liberated on fission of 1g 92235U

=200×106×1.6×1019×6.023×1023235=8.2×1010 Jg1

Total Energy produced in the reactor in 5 years

=1000×106×0.8×5×365×24×3600=1.261×1017 J

Mass of 92235U which underwent fission, m

=1.261×10178.2×1010

=1537.8 kg

The amount present initially in the reactor = 2m

=2 × 1537.8

=3075.6 kg

12. 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).

e++zAXZ1AY+v

Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

Answer:

For the electron capture, the reaction would be

ZAX+eZ1AY+ν+Q1

The mass defect and q value of the above reaction would be

Δm1=m(ZAX)+mem(Z1AY)Q1=([m(ZAX)m(Z1AY)]+me)c2

where m N (ZAX) and m N (Z1AY) are the nuclear masses of elements X and Y respectively

For positron emission, the reaction would be

ZAXZ1AY+e++ν¯+Q2

The mass defect and q value for the above reaction would be

Δm2=m(ZAX)m(Z1AY)meQ2=([m(ZAX)m(Z1AY)]me)c2

From the above values, we can see that if Q2 is positive, Q1 will also be positive but Q1 being positive does not imply that Q2 will also be positive.

Answer:

Let the abundances of 1225Mg and 1226Mg be x and y respectively.

x+y+78.99=100

y=21.01-x

The average atomic mass of Mg is 24.312 u

24.312=78.99×23.98504+x×24.98584+(100x)×25.98259100x9.3y=21.01xy=21.019.3y=11.71

The abundances of 1225Mg and 1226Mg are 9.3% and 11.71% respectively

14. (i) 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 2041Ca from the following data:

m(2040Ca)=39.962591u

m(2041Ca)=40.962278u

m(1326Al)=25.986895u

m(1327Al)=26.981541u

Answer:

The reaction showing the neutron separation is

2041Ca+E2040Ca+01n

E=(m(2040Ca)+m(01n)m(2041Ca))c2
E=(39.962591+1.00866540.962278)c2
E=(0.008978)u×c2

But 1u=931.5 MeV/c2

Therefore E=(0.008978) × 931.5

E=8.363007 MeV

Therefore to remove a neutron from the 2041Ca nucleus 8.363007 MeV of energy is required

14. (ii) 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 1327Al from the following data:

m(2040Ca)=39.962591u

m(2041Ca)=40.962278u

m(1326Al)=25.986895u

m(1327Al)=26.981541u

Answer:

The reaction showing the neutron separation is

1327Al+E1326Al+01n

E=(m(1326Ca)+m(01n)m(1327Ca))c2


E=(25.986895+1.00866526.981541)c2
E=(0.014019)u×c2

But 1u=931.5 MeV/c2

Therefore E=(0.014019) × 931.5

E=13.059 MeV

Therefore to remove a neutron from the 1327Al nucleus 13.059 MeV of energy is required

15. A source contains two phosphorous radio nuclides 1532P(T1/2=14.3d) and 1533P(T1/2=25.3d) . Initially, 10% of the decays come from 1533P . How long one must wait until 90% do so?

Answer:

Let initially there be N1 atoms of 1532P and N2 atoms of 1533P and let their decay constants be λ1 and λ2 respectively. Since initially the activity of 1533P is 1/9 times that of 1532P we have

N1λ1=N2λ29


Let after time t, the activity of 1533P be 9 times that of 1532P

N1λ1eλ1t=9N2λ2eλ2t


Dividing equation (ii) by (i) and taking the natural log of both sides, we get

λ1t=ln81λ2tt=ln81λ2λ1

where λ2=0.048/ day and λ1=0.027/ day
t comes out to be 208.5 days

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

88223Ra82209Pb+614C

88223Ra86219Rn+24He

Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer:

88223Ra82209Pb+614C

Δm=m(88223Ra)m(82209Pb)m(614C)=223.01850208.9810714.00324=0.03419u

1 u = 931.5 MeV/c2

Q=0.03419 × 931.5

=31.848 MeV

As the Q value is positive, the reaction is energetically allowed

88223Ra86219Rn+24He

Δm=m(88223Ra)m(86219Rn)m(24He)=223.01850219.009484.00260=0.00642u

1 u = 931.5 MeV/c2

Q=0.00642 × 931.5

=5.98 MeV

As the Q value is positive, the reaction is energetically allowed

17. Consider the fission of 92238U 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 58140Ce and 4499Ru . Calculate Q for this fission process. The relevant atomic and particle masses are

m(92238U)=238.05079u

m(58140Ce)=139.90543u

m(4499Ru)=98.90594u

Answer:

The fission reaction given in the question can be written as

92238U+01n58140Ce+4499Ru+10e

The mass defect for the above reaction would be

Δm=mN(92238U)+m(01n)mN(58140Ce)mN(4499Ce)10me

In the above equation, mN represents nuclear masses

Δm=m(92238U)92me+m(01n)m(58140Ce)+58mem(4499Ru)+44me10meΔm=m(92238U)+m(01n)m(58140Ce)m(4499Ru)Δm

Δm=238.05079+1.008665139.9054398.90594Δm=0.247995u

but 1u =931.5 MeV/c2

Q=0.247995 × 931.5

Q=231.007 MeV

Q value of the fission process is 231.007 MeV

18. (a)Consider the D–T reaction (deuterium-tritium fusion)

12H+13H24He+n Calculate the energy released in MeV in this reaction from the data:

m(12H)=2.014102u

m(13H)=3.016049u

Answer:

The mass defect of the reaction is

Δm=m(12H)+m(13H)m(24He)m(01n)
Δm=2.014102+3.0160494.0026031.008665
Δm=0.018883u

1u = 931.5 MeV/c2

Q=0.018883 × 931.5=17.59 MeV

18. (b) Consider the D–T reaction (deuterium-tritium fusion)

12H+13H24He+n

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.)

Answer:

To initiate the reaction, both nuclei would have to come into contact with each other.

Just before the reaction, the distance between their centres would be 4.0 fm.

The electrostatic potential energy of the system at that point would be

U=q24πϵ0d
U=9×109(1.6×1019)24×1015
U=5.76×1014J

The same amount of Kinetic Energy, K, would be required to overcome the electrostatic forces of repulsion to initiate the reaction

It is given that K=2×3kT2

Therefore, the temperature required to initiate the reaction is

T=K3k=5.76×10143×1.38×1023=1.39×109 K

19. Obtain the maximum kinetic energy of β - particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. You are given that

m(198Au)=197.968233u

m(198Hg)=197.966760u

1594197218194

Answer:

γ1 decays from 1.088 MeV to 0 V

Frequency of γ1 is

ν1=1.088×106×1.6×10196.62×1034
ν1=2.637×1020 Hz

Planck's constant, h=6.62 × 1034 Js,

E=hν

Similarly, we can calculate the frequencies of γ2 and γ3

ν2=9.988×1019 Hz
ν3=1.639×1020 Hz

The energy of the highest level would be equal to the energy released after the decay

Mass defect is

Δm=m(79196U)m(80196Hg)
Δm=197.968233197.966760
Δm=0.001473u

We know 1u = 931.5 MeV/c2

Q value= 0.001473 × 931.5=1.3721 MeV

The maximum Kinetic energy of β1 would be 1.3721-1.088=0.2841 MeV

The maximum Kinetic energy of β2 would be 1.3721-0.412=0.9601 MeV

20. 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.

Answer:

(a) 11H11H+11H+11H+11H24He

The above fusion reaction releases the energy of 26 MeV
The number of Hydrogen atoms in 1.0 kg of Hydrogen is 1000 NA
Therefore, 250 NA such reactions would take place
The energy released in the whole process is E1

=250×6.023×1023×26×106×1.6×1019=6.2639×1014J

(b) The energy released in the fission of one 92235U atom is 200 MeV

The number of 92235U atoms present in 1 kg of 92235U is N

N=1000×6.023×1023235N=2.562×1024


The energy released on the fission of N atoms is E2

E=2.562×1024×200×106×1.6×1019E=8.198×1013JE1E2=6.2639×10148.198×10138

21. 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.

Answer:

Let the amount of energy produced using nuclear power per year in 2020 be E

E=200000×106×0.1×365×24×36000.25 J

(Only 10% of the required electrical energy is to be produced by Nuclear power, and only 25% of therm-nuclear is successfully converted into electrical energy)

The amount of Uranium required to produce this much energy is M

=200000×106×0.1×365×24×3600×2350.25×200×106×1.6×1019×6.023×1023×1000 (N A =6.023 × 1023 , Atomic mass of Uranium is 235 g)

=3.076 × 104 kg

NCERT Solutions for Class 12 Physics Chapter 13 - Nuclei: Higher Order Thinking Skills (HOTS) Questions

Class 12 Physics Chapter 13 - Nuclei: Higher Order Thinking Skills (HOTS) Questions encourage students to apply concepts like nuclear binding energy, radioactivity, and nuclear reactions in complex scenarios. These advanced problems help in developing analytical skills and are very useful for competitive exam preparation.

Q1:

The energy released in the fusion of 2 kg of hydrogen deep in the sun is EH, and the energy released in the fission of 2 kg of 235U is EU. The ratio EHEU is approximately :
(Consider the fusion reaction as 411H+2e24He+2v+6γ+26.7MeV, energy released in the fission reaction of 235U is 200MeV per fission nucleus and NA=6.023×1023 )

Answer:

In each fusion reaction, 411H nuclei are used.

Energy released per Nuclei of 11H=26.74MeV

Energy released by 2 kg of hydrogen (EH )

=20001×NA×26.74MeV

Energy released by 2 kg Cranium (Ev)

=2000235×NA×200MeV

So,

EHEV=235×26.74×200=7.84

Approximately close to 7.62


Q2:

The disintegration energy Q for the nuclear fission of 235U140Ce+94Zr+n is _____ MeV.
Given atomic masses of
235U:235.0439u;140Ce;139.9054u,94Zr:93.9063u;n:1.0086u,

Value of c2=931MeV/u

Answer:

235U140Ce+94Zr+n

Disintegration energy
Q=(mRmp)c2 mR=235.0439ump=139.9054u+93.9063u+1.0086u=234.8203uQ=(235.0439u234.8203u)c2=0.2236c2=0.2236×931Q=208.1716


Q3:

The half-life of a radioactive isotope is 5.5 h. If there are initially 48×1232 atoms of this isotope, the number of atoms of the isotope remaining after 22 h is -

Answer:

Use, t1/2=tlog(12)[N(t)N0]

5.5=22log(12)x48×1032(12)x48×1032=110000x48×1032=14x=12×1032


Q4:

The kα radiation of M0(z=42) has a wavelength of 0.71 A˙, the wave length of the corresponding radiation of Cu(z=29)

Answer:

From Moseley's law for Kα - line, we have

1λα(z1)2λcuλm0=(zm01)2(zcu1)2=(41)2(28)2λm0=0.71Å,λcu=(0.71Å)×(41)2(28)2=1.52Å


Q5:

The count rate from 100 cm2 of a radioactive liquid is c. Some of the liquid is now discarded. The count rate of the remaining liquid is found to be (c10) after three half-lives. The volume of the remaining liquid in cm2 is:-

Answer:

Initial count rate (C R) for 1 cm3 of liquid =c100


After 3 half -lives, langle R for 1 cm3 of liquid

=18×c100


If u is the volume of remaining liquid, then

v×c800=c10v80=1v=80


Nuclei NCERT Solutions: Topics

Class 12 Physics Chapter 13 - Nuclei deals with the study of the atomic nucleus, its properties, and the forces that hold it together. The topics cover atomic masses, binding energy, nuclear force, radioactivity, and nuclear energy, which are crucial for understanding both natural processes and modern technology.

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Class 12 Physics Chapter 13 - Nuclei: Important Formulas

Class 12 Physics Chapter 13 - Nuclei formulas include key equations related to nuclear radius, binding energy, decay laws, half-life, and activity. These formulas are essential for solving numerical problems in board exams as well as JEE and NEET.

1. Atomic Masses and Composition of Nucleus

  • Nucleus = Protons + Neutrons (Nucleons)
  • Atomic number (Z), Mass number (A), Neutrons = A - Z
  • 1 atomic mass unit (u)=1.66×1027 kg
  • Isotopes: Same Z, different A

2. Size of the Nucleus

  • Radius: R=R0A1/3,R01.2×1015 m
  • Nuclear density is constant and very high ( 2.3×1017 kg/m3 )

3. Mass-Energy and Binding Energy

  • Mass defect: Actual mass < sum of nucleons
  • E=mc2 : Mass converts to energy
  • Binding energy = energy needed to break the nucleus
  • Higher binding energy = more stable nucleus

4. Nuclear Force

  • Strong, short-range, attractive force
  • Acts between all nucleons
  • Charge-independent and repulsive at very short distances

5. Radioactivity

  • Spontaneous decay of unstable nuclei
  • Types: Alpha (He nucleus), Beta (electron/positron), Gamma (photons)
  • Law: N(t)=N0eλt
  • Half-life: T1/2=0.693/λ

6. Nuclear Energy

  • Fission: Heavy nucleus splits (used in reactors)
  • Fusion: Light nuclei combine (in stars)
  • Both release huge energy due to binding energy changes

Approach to Solve Questions of Class 12 Physics Chapter 13 - Nuclei

In answering the questions of this chapter, attention should be drawn to developing the clarity of the stated problem, determining the correct physical principles, and systematically using formulas. Students are advised to be keen on derivations, units, and methods of approximation commonly examined in examinations. An organised approach to solving problems is not only time saving, but also minimises mistakes made when taking board as well as competitive exams, such as JEE as well as NEET.

  • Understand fundamental nuclear terms:

  1. Atomic number (Z), Mass number (A), Neutrons (N = A – Z)

  2. Nucleus structure: Protons and neutrons (nucleons)

  • Understand the nuclear size formula:
    Radius of nucleus: R=R0A1/3, where R01.2×1015 m
  • Apply mass-energy equivalence:
    E=mc2, where mass difference is utilised to generate energy
  • Calculate binding energy:
  1. Binding Energy (B.E) =Δm×931MeV
  2. Δm= mass defect = (total mass of nucleons – nucleus mass)
  • Understand the stability of nuclei:
  1. More B.E per nucleon ⇒ stable nucleus
  2. See the graph for binding energy per nucleon versus mass number
  • Learn about types of radioactive decay:

  1. Alpha decay: Mass number goes down by 4, atomic number goes down by 2
  2. Beta decay: Neutron → Proton (or vice versa), atomic number ±1
  3. Gamma decay: No change of mass or atomic number, only release of energy.
  • Apply the radioactive decay law:
  1. N=N0eλt, where λ is the decay constant
  2. Half-life: T1/2=0.693λ
  3. Activity: A=λN
  • Use appropriate units and convert them:
    Energy in MeV or joules, mass in u, time in seconds
  • Learn nuclear reactions:
    Balance the equations so that they retain mass number and atomic number.
  • Practice numerical + conceptual questions:

Emphasise half-life, decay, reaction equations, and energy calculations

What Extra Should Students Study Beyond NCERT for JEE/NEET?

Beyond the NCERT, students preparing for JEE/NEET should focus on advanced concepts of the Nuclei chapter, such as mass defect and binding energy curve analysis, nuclear reactions (fission & fusion), Q-value calculations, and detailed radioactive decay series. These topics strengthen problem-solving ability and are frequently tested in competitive exams.

NCERT Solutions for Class 12 Physics Chapter-wise

The NCERT Solutions for Class 12 Physics provide step-by-step, detailed answers to all chapters, helping students strengthen their conceptual understanding and problem-solving skills. These chapter-wise solutions are designed as per the latest CBSE curriculum and are equally useful for JEE, NEET, and other competitive exams. With structured explanations and solved examples, these solutions make learning easier and revision more effective.

Also Check:

NCERT solutions subject-wise

Also, check NCERT Exemplar Class 12 Solutions

Frequently Asked Questions (FAQs)

Q: Is the chapter Nuclei important for NEET and JEE Main?
A:

Yes the NCERT chapter Nuclei are important for both the exams. Both in NEET and JEE main syllabus the chapter Nuclei is present and 1 or 2 questions from the chapter can be expected for the exams. The questions discussed in the NCERT Solutions for the chapter Nuclei will give a better idea on how to use the formulas and give a better understanding of the concepts discussed.

Q: How nuclei class 12 ncert solutions is important for Board?
A:

NCERT solutions are important for the Board exam as they provide clear explanations, help in solving questions, cover all important topics, provide a structured approach to solving problems, and are designed with the exam pattern in mind, helping in exam-oriented preparation.

Q: What is the composition of the nucleus according to nuclei ncert solutions?
A:

The nucleus is made up of protons, which are positively charged particles, and neutrons, which are neutral particles.

Q: According to nuclei class 12 what is isotopes?
A:

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.

Q: What is the weightage of the chapter nuclei for CBSE board exam
A:

For CBSE board exam from NCERT class, 12 chapters 13 around 4 to 6 marks questions can be expected. All topics of the NCERT syllabus for the chapter Nuclei should be covered for the CBSE board exam.

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The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

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Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

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Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.