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NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

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Understanding Nuclei: Comprehensive NCERT Solutions for Class 12 Physics Chapter 13

CBSE board examiners always look for well-managed, stepwise detailed, accurate, and clear answers. Most of the questions are further extensions of basic NCERT questions; that’s why students look for mostly Chapter 13 which are accurate and solved with simplification. This chapter includes multiple numbers and makes the calculation process a little bit difficult. Thus, in this chapter, Chapter 13 with easy methods and less calculation are most important.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Physics

Chapter Name:

Chapter 13 - Nuclei

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

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Class 12 Physics Chapter 13 Chapter 13 has many NCERT questions and examples for problem-solving and practising. Thus, there are high chances that a number of new questions might be given in the examination, keeping these questions as the base.


Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

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Nuclei Chapter at a Glance - Class 12 NCERT Solutions

1.1 Isotopes

The atoms of an element, which have the same atomic number but different mass numbers, are called isotopes.


(i) 8O16, 8O17, 8O18

(ii) 17Cl35, 17Cl37

(iii) 82Pb206, 82Pb207, 82Pb208.


1.2 Isotones

The atoms whose nuclei have same number of neutrons are called isotones. 6C14, 7N15


1.3 Isobars

The atoms, which have same mass number but different atomic numbers, are called isobars.


(i) 1H3 and 1H3

(ii) 2Li7 and 4Be7

(iii) 28Ar40 and 29Ca40

(iv) 32Ge76 and 34Se76


1.4 Atomic Mass Unit

The atomic mass unit (a.m.u) is a very small unit of mass and it is found to be very convenient in nuclear physics.


Atomic mass unit is defined as 1/12th of the mass of one 6C12 atom.

According to Avogadro’s hypothesis, number of atoms in 12 g of 6C12 is equal to Avogadro number i.e. 6.023 × 1023.

Therefore, the mass of one carbon atom (6C12) is i.e. 1.992678 × 10–26 kg.

Therefore, 1 a.m.u. 

or 1 a.m.u. = 1.660565 × 10-27 kg


1.5 Energy Equivalent of Atomic Mass Unit

According to Einstein’s mass-energy equivalence relation, the energy equivalent of mass m is given by E = mc2

Where c is speed of light.

Suppose that m = 1 a.m.u = 1.660565 × 10-27 kg

Since, c = 2.998 × 108 ms-1, the energy equivalent of 1 a.m.u is given by 

1 a.m.u = (1.660565 × 10-27 kg) × (2.998 × 108 ms1)2
= 1.4925 × 10-10 J

Since, 1 MeV = 1.602 × 10-13 J, we have 

or 1 a.m.u = 931.5 MeV


1.6 Nuclear Size

The volume of the nucleus is directly proportional to the number of nucleons (mass number) constituting the nucleus. If R is the radius of the nucleus having mass number A, then

43πR3αA

or R α A1/3 or R = R0 A1/3       R0 = 1.2 × 10-15m


1.7 Nuclear Density

Mass of the nucleus of the atom of mass number A = A a.m.u = A × 1.660565 × 10-27 kg. If R is radius of the nucleus, then

Volume of nucleus = 43πR3=43(R03A1/3)3=43πR03A

Taking R0 = 1.2 × 10-15 m, we have

Density of the nucleus, ρ=mass of nucleusvolume of nucleus 

=A×1.66065×102743π(1.2×1015)3×A=2.26×1017kgm3

(independent of A)


NOTE:

  • The density of the nuclei of all the atoms is same as it is independent of mass number.

  • The high density of the nucleus (≈ 1017 kg m-3) suggests the compactness of the nucleus. Such examples of high densities are met in the form of neutron stars.


1.8 Mass Defect

The difference between the sum of the masses of the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect. It is denoted by Δm.


Let us calculate the mass defect in case of the nucleus of an atom zXA. The nucleus of the atom contains Z protons and (A−Z) neutrons. Therefore, if mN (zXA) is mass of the nucleus of the atom ZXA, then the mass defect is given by


Δm = [Zmp + (A−Z)mn −mN (zXA)]


The binding energy of a nucleus may be defined as the energy equivalent to the mass defect of the nucleus. It may be measured as the work required to be done to separate the nucleon an infinite distance apart, so that they no longer interact with each other.

If Δm is mass defect of a nucleus, then according to Einstein’s mass-energy relation, binding energy of the nucleus = Δm c2 (in joule).


Here, mass defect Δm has to be measured in kilogram.  In case, mass defect is measured in a.m.u., then


Binding energy of the nucleus = Δm × 931.5 (in MeV)

Binding energy = [Zmp + (A−Z)mn− mN (ZXA)] × 931.5 MeV


1.9 Binding Energy Per Nucleon

The binding energy per nucleon is the average energy required to extract one nucleon from the nucleus.


Thus, binding energy per nucleon =bindingenergyA


Chart

Description automatically generated with low confidence

Fig. 10.9


1.10 Packing Fraction

Packing fraction = (mass defect)/A.


1.11 Nuclear Fission

The process of splitting of a heavy nucleus into two nuclei of nearly comparable masses with liberation of energy is called nuclear fission.


92U235 + 0n1 → [92U236] → 56Ba141+ 36Kr92 + 30n1 + Q


Neutron reproduction factor is defined as the ratio of the rate of production of neutrons to the rate of loss of neutrons. Thus, k=rate of production of neutronsrate of loss of neutrons


A fission reaction will be steady, in case k = 1. In case k > 1, the fission reaction will accelerate and it will retard, in case  k < 1.


1.12 Nuclear Fusion

When two light nuclei fuse to form a larger nucleus, energy is released, this process is called nuclear fusion. Some examples of such energy liberating fusion reactions are:


11H+11H12H+e++v+0.42MeV

12H+12H23He+n+3.27MeV

12H+12H13H+11H+4.03MeV


In the first reaction, two protons combine to form a deuteron and a position with a release of 0.42 MeV energy. In second reaction, two deuterons combine to form the light isotope of helium. In third reaction, two deuterons combine to form a triton and a proton. For fusion to take place, the two nuclei must come close enough so that attractive short – range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is ~ 400 keV, and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough to overcome the coulomb barrier :


(32)kT=K 400 keV, which gives T ~ 3 × 109 K. 


When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion.


2.1 Laws of Radioactivity Decay

Rutherford and Soddy studied the phenomenon of radioactivity in details and formulated the following laws, known as the laws of radioactive decay:


  • Radioactivity is a spontaneous phenomenon and one cannot predict, when a particular atom in a given radioactive sample will undergo disintegration.

  • When a radioactive atom disintegrates, either an α-particle (nucleus of helium) or a β-particle (electron) is emitted.

  • The emission of an α-particle by a radioactive atom result in a daughter atom, whose atomic number is 2 units less and mass number is 4 units less than that of the parent atom.

zXAαdecayz2YA4

  • The emission of a β-particle by a radioactive atom result in a daughter atom, whose atomic number is 1 unit more but mass number is same as that of the parent atom.

zXAβdecayz+1YA

  • The number of atoms disintegrating per second of a radioactive sample at any time is directly proportional to the number of atoms present at that time.  The rate of disintegration of the sample cannot be altered by changing the external factors, such as pressure, temperature etc.  It is known as radioactive decay law.


According to radioactive decay law, the rate of disintegration at any time t is directly proportional to the number of atoms present at time t i.e.


dNdtNordNdt=λN.

Where the constant of proportionally λ is called decay constant of the radioactive sample.  It is also known as disintegration constant or transformation constant.  Its value depends upon the nature of the radioactive sample.  Further, the negative sign indicates that the number of the atoms of the sample decreases with the passage of time.


From equation, we have dNdt=λdt.

Integrating N0NdNN=0tλdt

Or logeNN0=λt

Or NN0=eλt

Or N = N0e−λt


2.3 Half Life

Consider that a radioactive sample contains N0 atoms at time t = 0.  Then, the number of atoms left behind after time t is given by N = N0 e−λ t

From the definition of half-life, it follows that when t = t1/2, N=N02.

Setting the above condition in equation, we have N02=N0eλt1/2

Or eλt1/2=1/2 Or e^{\lambda t_{1/2}} = 2

Or  λT = loge 2 = 2.303 log10 2 = 2.303 × 0.3010 = 0.693

Or t1/2=0.693λ

Thus, half life of a radioactive substance is inversely proportional to its decay constant and is a characteristic property of its nucleus. It cannot be altered by any known method.


2.4 Mean life or Average Life

The average life of a radioactive substance is defined as the average time for which the nuclei of the atoms of the radioactive substance exist. It is defined by tavg.


tavg=1λ

Mastering Class 12 Physics Chapter 13: Nuclei - MCQs, Question and Answers, and Tips for Success

1. Questions and Answers

(a) Two stable isotopes of lithium 36Liand 37Li have respective abundances of 7.5 and 92.5These isotopes have masses 6.01512uand7.01600u, respectively. Find the atomic mass of lithium.

Ans: We are given the following information:

Mass of 36Li lithium isotope, m1=6.01512u

Mass of 37Li lithium isotope, m2=7.01600u

Abundance of36Li, n1=7.5

Abundance of  37Li, n2=92.5

The atomic mass of lithium atom is given by,

m=m1n1+m2n2n1+n2

Substituting the given values, we get,

m=6.01512×7.5+7.01600×92.592.5+7.5

m=6.940934u

Therefore, we found the atomic mass of lithium atoms to be 6.940934u


(b) Boron has Two Stable Isotopes 510B and511B. Their respective masses are 10.01294uand11.00931u, and the atomic mass of boron is 10.811u. Find the abundances of 510B and 511B .

Ans: We are given:

Mass of 510B Boron isotope, m1=10.01294u

Mass of 511B lithium isotope, m2=11.00931u

Abundance of510B, n1=x

Abundance of  511B, n2=(100x)

We know the atomic mass of boron to be, m=10.811u

The atomic mass of lithium atom is given by,

m=m1n1+m2n2n1+n2

Substituting the given values, we get,

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

1081.11=10.01294x+1100.93111.00931x

x=19.8210.99637=19.89

And, 100x=80.11

Therefore, we found the abundance of 510Band 511Bto be 19.89 and 80.11respectively. 


2. The three stable isotopes of neon: 1020Ne, 1021Ne and 1022Ne have respective abundances of 90.51The atomic masses of the three isotopes are19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Ans: We are given that:

Atomic mass of  1020Ne, m1=19.99u

Abundance of 1020Ne, η1=90.51

Atomic mass of1021Ne, m2=20.99u

Abundance of 1021Ne, η2=0.27

Atomic mass of 1022Ne, m3=21.99u

Abundance of 1022Ne, η3=9.22

The average atomic mass of neon could be given as,

m=m1η1+m2η2+m3η3η1+η2+η3

Substituting the given values, we get, 

m=19.99×90.51+20.99×0.27+21.99×9.2290.51+0.27+9.22

m=20.1771u

The average atomic mass of neon is thus found to be 20.177u.


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

Ans: We are given:

Atomic mass of nitrogen (7N14), m=14.00307u

A nucleus of 7N14nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus would be, Δm=7mH+7mnm

Where,

Mass of a proton, mH=1.007825u

Mass of a neutron, mn=1.008665u

Substituting these values into the above equation, we get, 

Δm=7×1.007825+7×1.00866514.00307

Δm=7.054775+7.0605514.00307

Δm=0.11236u

But we know that, 1u=931.5MeV/c2

Δm=0.11236×931.5MeV/c2

Now, we could give the binding energy as, 

Eb=Δmc2

Where, c=speed of light =3×108ms2

Now, Eb=0.11236×931.5(MeVc2)×c2

Eb=104.66334MeV

Therefore, we found the binding energy of a Nitrogen nucleus to be 104.66334MeV


4. Obtain the binding energy of the nuclei 2656Fe and 83209Bi in units of MeV from the following data: m(2656Fe)=55.934939u, m(83209Bi)=208.980388u

Ans: We are given the following:

Atomic mass of 2656Fe, m1=55.934939u

2656Fe nucleus has 26 protons and 5626=30neutrons

Hence, the mass defect of the nucleus would be, Δm=26×mH+30×mnm1

Where, Mass of a proton,mH=1.007825u

Mass of a neutron,mn=1.008665u

Substituting these values into the above equation, we get, 

Δm=26×1.007825+30×1.00866555.934939

Δm=26.20345+30.2599555.934939

Δm=0.528461u

But we have, 1u=931.5MeV/c2

Δm=0.528461×931.5MeV/c2

The binding energy of this nucleus could be given as,

Eb1=Δmc2

Where, c = Speed of light

Eb1=0.528461×931.5(MeVc2)×c2

Eb1=492.26MeV

Now, we have the average binding energy per nucleon to be, 

B.E=492.2656=8.79MeV

Also, atomic mass of 83209Bi, m2=208.980388u

We know that, 83209Bi nucleus has 83 protons and 20983=126 neutrons

Where,

Mass of a proton, mH=1.007825u

Mass of a neutron, mn=1.008665u

Δm=83×1.007825+126×1.008665208.980388

Δm=83.649475+127.091790208.980388

Δm=1.760877u

But we know, 1u=931.5MeV/c2

Hence, the binding energy of this nucleus could be given as,

Eb2=Δmc2=1.760877×931.5(MeVc2)×c2

Eb2=1640.26MeV

Average binding energy per nucleon is found to be =1640.26209=7.848MeV

Hence, the average binding energy per nucleon is found to be 7.848MeV


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 2963Cu atoms (of mass62.92960u).

Ans: We are given:

Mass of a copper coin, m=3g

Atomic mass of 29Cu63atom, m=62.92960u

The total number of 2963Cu atoms in the coin, N=NA×mMass number

Where,NA= Avogadro number=6.023× 1023atoms/g

Mass number = 63g

N=6.023×1023×363=2.868×1022

29Cu63nucleus has 29 protons and (6329)=34neutrons

Mass defect of this nucleus would be, Δm=29×mH+34×mnm

Where,

Mass of a proton, mH=1.007825u

Mass of a neutron, mn=1.008665uΔm=29×1.007825+34×1.00866562.9296=0.591935u

Mass defect of all the atoms present in the coin would be, 

Δm=0.591935×2.868×1022=1.69766958×1022u

But we have, 1u=931.5MeV/c2

Δm=1.69766958×1022×931.5MeV/c2

Hence, the binding energy of the nuclei of the coin could be given as:

Eb=Δmc2=1.69766958×1022×931.5(MeVc2)×c2

Eb=1.581×1025MeV

But, 1MeV=1.6×1013J

Eb=1.581×1025×1.6×1013

Eb=2.5296×1012J

This much energy is needed to separate all the neutrons and protons from the given coin.

6. Write the Nuclear Reactions For:

(a) αdecay of 88226Ra

Ans: We know that, αis basically a nucleus of Helium (2He4)and βis an electron (e for β and e+ for β+). In every α-decay, there is a loss of 2 protons and 2 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

For the given case, the nuclear reaction would be, 

88Ra22686Rn222+2He4


(b) αdecay of 94242Pu

Ans: 94242Pu92238U+24He

  1. βdecay of 1532P

Ans: 1532P1632S+e+ν¯

  1. βdecay of 83210Bi

Ans: 83210B84210Po+e+ν¯

  1. β+decay of 611C

Ans: 611C511B+e++ν

  1. β+decay of 4397Tc

Ans: 4397Tc4297Mo+e++ν

  1. Electron capture of 54120Xe

Ans: 54120Xe+e+53120I+ν


7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to:

  1. 3.125% of Its Original Value? 

Ans: We are said that, Half-life of the radioactive isotope= T years

Original amount of the radioactive isotope =N0

(a) After decay, let the amount of the radioactive isotope be N. 

It is given that only 3.125 of N0remains after decay. Hence, we could write,

NN0=3.125

But we know that, NN0=eλt

Where, λ=Decay constant and t=Time

λt=132

λt=lnlln32

λt=03.4657

t=3.4657λ

But, since λ=0.693T

t=3.466(0.693T)

t5T years

Therefore, we found that the isotope will take about 5T years in order to reduce to 3.125 of its original value. 

  1. 1% of its original value?

Ans: After decay, let the amount of the radioactive isotope be N

It is given that only 1% of N0 remains after decay. Hence, we could write:

NN0=1

But we know, NN0=eλt

eλt=1100

λt=ln1ln100

λt=04.602

t=4.6052λ

Since we have, λ=0.639T

t=4.6052(0.693T)

t=6.645T years

Therefore, we found that the given isotope would take about 6.645T years to reduce to 1% of its original value. 


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 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 civilisation.

Ans: We are given that: 

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 614C, T12=5730 years

The decay rate of the specimen obtained from the Mohenjodaro site:

R=9 decays/min

Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant, λ and time, t as:

NN=RR=eλt

eλt=915=35

λt=loge35=0.5108

t=0.5108λ

But we know, 

λ=0.693T12=0.6935730

t=0.5108(0.6935730)=4223.5 years

Therefore, the approximate age of the Indus-Valley civilization is found to be 4223.5 years.


9. Obtain the Amount of 2760Co necessary to provide a radioactive source of 8.0 mCistrength. The half-life of 2760Co is 5.3 years.

Ans: We know that, 

The strength of the radioactive source could be given as, 

dNdt=8.0mCi

dNdt=8×103×3.7×1010=29.6×107decay/s

Where, N is the required number of atoms. 

Half life of 2760Co, T12=5.3years

T12=5.3×365×24×60×60=1.67×108s

For decay constant λ, we could give the rate of decay as, 

dNdt=λN

Where, λ=0.693T12=0.6931.67×108s1

N=1λdNdt=29.6×107(0.6931.67×108)=7.133×1016atoms

Now for 27Co60, Mass of Avogadro number of atoms =60g

Then, mass of 7.133×1016atoms=60×7.133×10166.023×1023=7.106×106g

Therefore, the amount of 27Co60that is required for the purpose is 7.106×106g


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

Ans: We know that, 

Half life of 3890Sr, t12=28years=28×365×24×3600=8.83×108s

Mass of the isotope, m=15mg

90g of 3890Sr atom contains Avogadro number of atoms. So, 15mg of 3890Sr contains, 

6.023×1023×15×10390=1.0038×1020number of atoms

Rate of disintegration would be, dNdt=λN

Where, λis the decay constant given by, λ=0.6938.83×108s1

dNdt=0.693×1.0038×10208.83×108=7.878×1010atoms/s

Therefore, we found the disintegration rate of 15mg of given isotope to be 7.878×1010atoms/s.


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

Ans: We know that, 

Nuclear radius of the gold isotope 79Au197=RAu

Nuclear radius of the silver isotope 47Ag107=RAg

Mass number of gold, AAu=197

Mass number of silver, AAg=107

We also know that the ratio of the radii of the two nuclei is related with their mass numbers as:

RAuRAg=(AAuAAg)13=1.2256

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


12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of: [Given:m(88226Ra)=226.02540u, m(89222Rn)=222.01750u,  m(86220Rn)=220.01137u, m(84216Po)=216.00189u ]

  1. 88226Ra

Ans: We know that, 

Alpha particle decay of 8826Ra emits a helium nucleus. As a result, its mass number reduces to 222=(2264)and its atomic number reduces to86=(882). This is shown in the following nuclear reaction:

88226Ra86222Ra+24He

Qvalue of emitted αparticle = (Sum of initial mass  Sum of final mass)c2Where, c = Speed of light 

It is also given that:

m(88226Ra)=226.02540u

m(86220Rn)=220.01137u

m(24He)=4.002603u

On substituting these values into the above equation, 

Q value =[226.02540(222.01750+4.002603)]uc2

Q value=0.005297uc2

But we know, 1u=931.5MeV/c2

Q=0.005297×931.54.94MeV

Kinetic energy of the αparticle=(Mass number after decayMass number before decay)×Q

K.Eα=222226×4.94=4.85MeV

Hence, the Kinetic energy of the alpha particle is found to be 4.85MeV.


  1. 86220Rn

Ans: We know that, Alpha particle decay of 86220Rn could be given as, 

86220Rn84216Po+24He

We are also given, 

Mass of 86220Rn=220.01137u

Mass of 84216Po=216.00189u

Now, Q value could be given as, 

Qvalue=[220.01137(216.00189+4.00260)]×931.5641MeV

Now, we have the kinetic energy as, 

K.Eα=(2204220)×6.41=6.29MeV

The kinetic energy of the alpha particle is found to be 6.29MeV


13. The Radionuclide 611C decays according to,

611C511B+e++ν ; T12=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. 

Ans: The given nuclear reaction is, 

611C511B+e++ν

Half life of 611C nuclei, T12=20.3min

Atomic masses are given to be:

m(611C)=11.011434u

m(611B)=11.009305u

Maximum energy that is possessed by the emitted positron would be 0.960MeV. The change in the Q - value (ΔQ) of the nuclear masses of the 611C

ΔQ=[m(6C11)[m(511B)+me]]c2…………………….. (1)

Where,me=Mass of an electron or positron = 0.000548u

c= Speed of light

m=Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we will have to add 6me in the case of 11C and 5mein case of 11B.

Hence, equation (1) would now reduce to,

ΔQ=[m(6C11)m(511B)2me]c2

Where, m(6C11) and m(511B)are the atomic masses. 

Now, we have the change in Q value as, 

ΔQ=[11.01143411.0093052×0.000548]c2=(0.001033c2)u

But we know, 1u=931.5MeV/c2

ΔQ=0.001033×931.50.962MeV

We see that the Q value is almost comparable to the maximum energy of the emitted positron. 


14. The nucleus  1023Ne decays by βemission. Write down the βdecay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m(1023Ne)=22.994466u

m(1123Na)=22.989770u

Ans: We know that: In βemission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. βemission of the nucleus could be given by, 

1023Ne1123Na+e+ν¯+Q

It is also given that:

Atomic mass of 1023Ne=22.994466u

Atomic mass of 1123Na=22.989770u

Mass of an electron, me=0.000548u

Q value of the given reaction could be given as:

Q=[m(1023Ne)[m(1123Na)+me]]c2

There are 10 electrons in 10Ne23and 11 electrons in 1123Na. Hence, the mass of the electron is cancelled in the Q-value equation.

Q=[22.99446622.9897770]c2=(0.004696c2)u

But we have, 1u=931.5MeV/c2

Q=0.004696×931.5=4.374MeV

The daughter nucleus is too heavy as compared to that of e and ν¯. Hence, it carries negligible energy. The kinetic energy of the antineutrino is found to be nearly zero. 

Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e.,4.374MeV.


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.

Atomic masses are given to be: m(12H)=2.014102u, m(13H)=3.016049u,

m(612C)=12.000000u, m(1020Ne)=19.992439u

  1. 11H+13H12H+12H

Ans:The given nuclear reaction is:

11H+13H12H+12H

Atomic mass of 11H=1.007825u

Atomic mass of 13H=3.0164049u

Atomic mass of 12H=2.014102u

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

Q=[m(11H)+m(13H)2m(12H)]c2

Q=[1.007825+3.0160492×2.014102]c2=(0.00433c2)u

But we know, 1u=931.5MeV/c2

Q=0.00433×931.5=4.0334MeV

The negative Q-value of this reaction shows that the given reaction is endothermic. 


  1. 612C+612C1020Ne+24He

Ans: We are given that, 

Atomic mass of 612C=12.0u

Atomic mass of 1012Ne=19.992439u

Atomic mass of 24He=4.002603u

The Q-value here could be given as, 

Q=[2m(612C)m(1020Ne)m(24He)]c2

Q=[2×12.019.9924394.002603]c2=(0.004958c2)u=0.004958×931.5

Q=4.618377MeV

Since the Q-value is found to be positive, the reaction could be considered exothermic. 


16. Suppose, we think of fission of a 2656Fe nucleus into two equal fragments of 13Al28. Is fission energetically possible? Argue by working out Q of the process. Given: m(2656Fe)=55.93494u and m(1328Al)=27.98191u

Ans: We know that the fission of 2656Fe could be given as, 

2656Fe21328Al

We are also given, atomic masses of 2656Fe and 1328Al as 55.93494u and 27.98191u respectively. 

The Q-value here would be given as, 

Q=[m(2656Fe)2m(1328Al)]c2

Q=[55.934942×27.98191]c2=(0.02888c2)u

But, 1u=931.5MeV/c2

Q=0.02888×931.5=26.902MeV

The Q value is found to be negative and hence we could say that the fission is not possible energetically. In order for a reaction to be energetically possible, the Q-value must be positive. 


17. The fission properties of  94239Pu are very similar to those of 92235U .The average energy released per fission is180MeV. How much energy, in MeV, is released if all the atoms in 1kgof pure 94239Pu undergo fission?

Ans: We are given that the average energy released per fission of 94239Pu, Eav=180MeV

The amount of pure 94Pu239, m=1kg=1000g

Avogadro number, NA=6.023×1023

Mass number of 94239Pu=239g

1 mole of 94Pu239contains Avogadro number of atoms.

1g of 94Pu239 contains (NAmass number×m) atoms

(6.023×1023239×1000)=2.52×1024atoms

Total energy released during the fission of 1kg of 94239Pucould be calculated as:

E=Eav×2.52×1024=180×2.52×1024=4.536×1026MeV

Therefore, 4.536×1026MeV is released if all the atoms in 1kg of pure 94Pu239undergo fission.


18. A 1000MW fission reactor consumes half of its fuel in 5.00 y. How much 

92235U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92235U and that this nuclide is consumed only by the fission process.

Ans: We are said that the half life of the fuel of the fission reactor, t12=5years

t12= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1g of 92235U nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235g of  92235U contains 6.023 × 1023atoms.

1g of 92235U contains 6.023×1023234atoms

The total energy generated per gram of 92235U is calculated as:

E=6.023×1023235×200MeV/g=200×6.023×1023×1.6×1019×106235=8.20×1010J/g

The reactor operator operates only 80% of the time. Therefore, the amount of 92235U consumed in 5years by the 1000MW fission reactor could be calculated as, 

5×80×60×60×365×24×1000×106100×8.20×1010g1538kg

So, the initial amount of 92235U=2×1538=3076kg

Hence, we found the initial amount of uranium to be 3076kg.


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

12H+12H23He+n+3.27MeV

Ans: The fusion reaction is given to be:

12H+12H23He+n+3.27MeV

Amount of deuterium, m=2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 1023atoms.

2.0 kg of deuterium contains 6.023×10232×2000=6.023×1026atoms atoms

It could be inferred from the given reaction that when two atoms of deuterium fuse, 3.27MeVenergy is released.

Therefore, the total energy per nucleus released in the fusion reaction would be:

E=3.272×6.023×1026MeV=3.272×6.023×1026×1.6×1019×106

E=1.576×1014J

Power of the electric lamp is given to be, P=100W=100J/s, that is, the energy consumed by the lamp per second is 100J. 

Now, the total time for which the electric lamp glows could be calculated as, 

t=1.576×1014100=1.576×1014100×60×60×24×365

t4.9×104years

Hence, the total time for which the electric lamp glows is found to be 4.9×104years.


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 2.0fm.)

Ans: When two deuterons collide head-on, the distance between their centres, d could be given as:

Radius of 1st deuteron +Radius of 2nd deuteron

Radius of a deuteron nucleus=2fm=2×1015m

d=2×1015+2×1015=4×1015m

Also, charge on a deuteron = Charge on an electron =e=1.6×1019C

Potential energy of the two-deuteron system could be given by, 

V=e24πε0d

Where, ε0is the permittivity of free space. 

Also, 14πε0=9×109Nm2C2

V=9×109×(1.6×1019)24×1015J=9×109×(1.6×1019)24×1015×(1.6×1019)eV

V=360keV

Therefore, we found the height of the potential barrier of the two-deuteron system to be 360keV. 


21. From the relation R=R0A13, 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 know the expression for nuclear radius to be:

R=R0A13

Where, R0 is a Constant and A is the mass number of the nucleus

Nuclear matter density would be,

ρ=Mass of the nucleusVolume of the nucleus

Now, let m be the average mass of the nucleus, then, mass of the nucleus =mA

Nuclear density,

ρ=mA43πR3=3mA4π(R0A13)3=3mA4πR03A

ρ=3m4πR03

Therefore, we found the nuclear matter density to be independent of A and it is found to be nearly constant. 


22. For the β+(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K− shell, is captured by the nucleus and a neutrino is emitted).

e++ZAXZ1AY+ν

Show that if  β+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 could be written as:

e++ZAXZ1AY+ν+Q1………………………… (1)

Let the amount of energy released during the positron capture process be Q2. The nuclear reaction could be written as:

ZAXZ1AY+e++ν+Q2………………………… (2)

Let, mN(ZAX) be the nuclear mass of ZAX, mN(Z1AY) be the nuclear mass of Z1AY

m(ZAX) be the atomic mass of ZAX

m(Z1AY) be the nuclear mass of Z1AY

me be the mass of an electron, c be the speed of light, then, the Q-value of the electron capture reaction could be given as, 

Q1=[mN(ZAX)+memN(Z1AY)]c2

Q1=[m(ZAX)Zme+mem(Z1AY)+(Z1)me]c2

Q1=[m(ZAX)m(Z1AY)]c2………………… (3)

The Q-value of the positron capture reaction could be given as, 

Q2=[mN(ZAX)mN(Z1AY)me]c2

Q1=[m(ZAX)Zmem(Z1AY)+(Z1)meme]c2

Q1=[m(ZAX)m(Z1AY)2me]c2………………… (4) 

It can be inferred that if Q2>0, then; Also, if Q1>0, it does not necessarily mean that Q2>0. In other words, we could say that if β+emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is so because the Q-value must be positive for an energetically-allowed nuclear reaction.


23. In a periodic table the average atomic mass of magnesium is given as 24.312u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are:

1224Mg(23.98504u), 1225Mg(24.98584u) and 1226Mg(25.98259u)

The natural abundance of 12Mg24is 78.99% by mass. Calculate the abundances of the other two isotopes.

Ans: We are given:

Average atomic mass of magnesium, m=24.312 u

Mass of magnesium 1224Mg isotope, m1=23.98504 u

Mass of magnesium 1225Mg isotope, m2=24.98584 u

Mass of magnesium 1226Mg isotope, m3=25.98259 u

Abundance of 1224Mg, η1=78.99 

Abundance of  1225Mg, η2=x

Now, the abundance of 1226Mg, η3=100x78.99

Also, we have the relation for the average atomic mass as:

m=m1η1+m2η2+m3η3η1+η2+η3

24.312=23.98504×78.99+24.98584×x+25.98259×(21.01x)100

0.99675x=9.2725255

x9.3

And, 21.01x=11.71

Therefore, we found the abundance of 1225Mg to be 9.3% and that of 1226Mg to be 11.71%.


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 2041Ca and 1327Al from the following data:

m(2040Ca)=39.962591u, m(2041Ca)=40.962278u, m(1326Al)=25.986995u, 

m(1327Al)=26.981541u

Ans: For a neutron removal from 20Ca41nucleus, the corresponding nuclear reaction could be written as, 

2041Ca2040Ca+01n

We are given:

m(2040Ca)=39.962591u

m(2041Ca)=40.962278u

m(0n1)=1.008665u

Now, the mass defect for this reaction could be given by, 

Δm=m(2040Ca)+(01n)m(2041Ca)

Δm=39.962591+1.00866540.962278=0.008978u

But we know, 1u=931.5MeV/c2

Δm=0.008978×931.5MeV/c2

Now, we could calculate the energy required for the neutron removal by, 

E=Δmc2

E=0.008978×931.5=8.363007MeV

For the case of 1327Al, the neutron removal reaction could be written as, 

1327Al1326Al+01n

We are given,

m(1326Al)=25.986995u

m(1327Al)=26.981541u

Now, the mass defect here could be given by, 

Δm=m(1326Al)+m(01n)m(1327Al)

Δm=25.986895+1.00866526.981541=0.014019u

Δm=0.014019×931.5MeV/c2

Therefore, the energy that is required for the removal of neutron would be, 

E=Δmc2=0.014019×931.5

E=13.059MeV


25. A source contains two phosphorous radio nuclides 1532P (T12=14.3d) and 1533P (T12=25.3d)Initially, 10% of the decays come from 1533P. How long must one wait until 90% do so?

Ans: We are given:

Half life of  1532P (T12=14.3d)

Half life of  1533P (T12=25.3d)

Now, we know that nucleus decay is 10% of the total amount of decay.

Also, the source has initially 10% of 1532P nucleus and 90% of 1532P nucleus.

Suppose after t days, the source has 10% of 1532P nucleus and 90% of 1533P 

nucleus.

Initially we have:

Number of 1533P nucleus =N

Number of 1532P nucleus =9N

Finally:

Number of 1533P nucleus =9N

Number of 1532P nucleus=N

For 1532P nucleus, we could write the number ratio as:

N9N=(12)tT12

N=9N(2)t14.3………………………. (1)

Now, for 1533P, we could write the number ratio as, 

9NN=(12)1T12

9N=N(2)t25.3…………………………. (2)

We could now divide equation (1) by equation (2) to get, 

19=9×2(t25.3t14.3)

181=2(11t25.3×14.3)

log1log81=11t25.3×14.3log1

11t25.3×14.3=01.9080.301

t=25.3×14.3×1.90811×0.301208.5days

Therefore, we found that it would take about 208.5days for 90% decay of 1533P.


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

88223Ra82209Pb+614C

88223Ra86219Rn+24He

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

Ans: Consider a 614C emission nuclear reaction, 

88223Ra82209Pb+614C

We know that:

Mass of 88223Ra, m1=223.01850u

Mass of 614C, m3=14.00324u

Now, the Q-value of the reaction could be given as:

Q=(m1m2m3)c2

Q=(223.01850208.9810714.00324)c2=(0.03419c2)u

But we have, 1u=931.5MeV/c2

Q=0.03419×931.5

Q=31.848MeV

Hence, the Q-value of the nuclear reaction is found to be 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now consider a 24He emission nuclear reaction:

88223Ra86229Rn+24He

We know that:

Mass of 88223Ra, m1=223.01850

Mass of 82219Rn, m2=219.00948

Mass of 24He, m3=4.00260

Q-value of this nuclear reaction could be given as:

Q=(m1m2m3)c2

Q=(223.01850219.009484.00260)c2

Q=(0.00642c2)u

Q=0.00642×931.5=5.98MeV

Therefore, the Q-value of the second nuclear reaction is found to be 5.98MeV. Since the value is positive, we could say that the reaction is energetically allowed.

27. 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

Ans: We are given:

In the fission of 92238U, 10 βparticles decay from the parent nucleus. The nuclear reaction can be written as:

92238U+01n58140Ce+4499Ru+1010e

It is also given that:

Mass of a nucleus of 92238U, m1=238.05079u

Mass of a nucleus of 58140Ce, m2=139.90543u

Mass of nucleus of 4499Ru, m3=98.90594u

Mass of a neutron 01n, m4=1.008665u

Q-value of the above equation would be,

Q=[m(92238U)+m(01n)m(58140Ce)m(4499Ru)10me]c2

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

m(92238U)=m192me

m(58140Ce)=m258me

m(4499Ru)=m344me

m(01n)=m4

Q=[m192me+m4m2+58mem3+44me10me]c2

Q=[m1+m4m2m3]c2=[238.0507+1.008665139.9054398.90594]c2

Q=[0.247995c2]u

But 1u=931.5MeV/c2

Q=0.247995×931.5 =231.007MeV

Therefore, the Q-value of the fission process is found to be 231.007MeV.


28. Consider the D−T reaction (deuterium−tritium fusion) 

12H+13H24He+n

  1. Calculate the energy released in MeV in this reaction from the data: m(12H)=2.014102u , m(13H)=3.016049u

Ans: Consider the D-T nuclear reaction, 

12H+13H24He+n

We are also given that:

Mass of 12H, m1=2.014102u

Mass of 13H, m2=3.016049u

Mass of 24He, m3=4.002603u

Mass of 01n, m4=1.008665u

Now, the Q-value of the given D-T reaction would be:

Q=[m1+m2m3m4]c2

Q=[2.014102+3.0160494.0026031.008665]c2

Q=[0.018883c2]u

But 1u=931.5MeV/c2

Q=0.018883×931.5=17.59MeV


  1. Consider the radius of both deuterium and tritium to be approximately 2.0fm. 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(3kT2); k = Boltzmann’s constant, T = absolute temperature.)

Ans: We are given:

Radius of deuterium and tritium, r2.0fm=2×1015m 

Distance between the two nuclei at the moment when they touch each other,

d=r+r=4×1015m

Charge on the deuterium nucleus =e

Charge on the tritium nucleus =e

Hence, the repulsive potential energy between the two nuclei could be given as:

V=e24πε0d

Where, ε0=Permittivity of free space

14πε0=9×109Nm2c2

V=9×109×(1.6×1019)24×1015=5.76×1014J=5.76×10141,6×1019

V=3.6×105eV=360keV

Therefore, 5.76×1014J or  360keVof kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.

However, we are also given that:

KE=2×3kT2

Where, k=Boltzmann constant 

T=Temperature required for triggering the reaction

T=KE3k=5.76×10143×1.38×1023=1.39×109K

Therefore, we found that the gas must be heated to a temperature of 1.39×109K to initiate the reaction.


29. Obtain the maximum kinetic energy of βparticles, and the radiation frequencies of γ decays in the decay scheme shown in figure. You are given that:

m(198Au)=197.968233u

m(198Hg)=197.966760u

(Image will be Uploaded Soon)

Ans: It can be observed from the given γ-decay diagram that γ1decays from the 1.088MeV energy level to the 0MeV energy level. Hence, the energy corresponding to γ1-decay is given as:

E1=1.0888=1.088MeV

hν1=1.088×1.6×1019×106J 

Where, Planck's constant  h=6.6×1034Js

ν1=Frequency of radiation radiated by γ1decay

ν1=E1h

ν1=1.088×1.6×1019×1066.6×1034=2.637×1020Hz

It can be observed from the given γdecay diagram that γ2decays from the 0.412MeVenergy level to the 0MeVenergy level.

Now, the energy corresponding to γ2-decay could be given as:

E2=0.4120=0.412MeV

hν2=0.412×1.6×1019×106J

Where, ν2=Frequency of radiation radiated by γ2decay

ν2=E2h=0.412×1.6×1019×1066.6×1034=9.988×1019Hz

It can be observed from the given γ-decay diagram that γ3-decays from the 1.088MeVenergy level to the 0.412MeVenergy level.

Now, the energy corresponding to γ3 -decay is given as:

E3=1.0880.412=0.676MeV

hν3=0.676×1019×106

Where, ν3=Frequency of radiation radiated by γ3decay

ν3=E3h=0.676×1.6×1019×1066.6×1034=1.639×1020Hz

Mass of m(78198Au) =197.968233u

Mass of m(80198Hg)=197.966760u

1u=931.5MeV/c2

Energy of the highest level could be given as:

E=[m(78198Au)m(80190Hg)]=197.968233197.966760=0.001473u

E=0.001473×931.5=1.3720995MeV

β1decays from the 1.3720995MeVlevel to the 1.088MeVlevel

Maximum kinetic energy of the β1particle =1.3720995  1.088

K.E=0.2840995MeV

β2 decays from the 1.3720995MeVlevel to that of the 0.412MeVlevel. Now, we find the maximum kinetic energy of the β2particle to be, 

K.Emax=1.37209950.412=0.9600995MeV

Therefore, we found the maximum kinetic energy of the β2particle to be 0.9600995MeV


30. Calculate and Compare the Energy Released By

  1. fusion of 1.0kg of hydrogen deep within Sun and

Ans: We are given:

Amount of hydrogen, m=1 kg=1000g

1 mole, i.e., 1g of hydrogen (11H) contains 6.023×1023atoms.

That is, 1000g of 11H contains 6.023×1023atoms.

Within the sun, four 11H nuclei combine and form one 24He nucleus. In this process 26MeVof energy is released.

Hence, the energy released from the fusion of 1 kg 11H is:

E1=6.023×1023×26×1034=39.1495×1026MeV

Therefore, we found the energy released during the fusion of 1kg 11H is:

E1=6.023×1023×26×1034=39.1495×1026MeV

Hence, the energy released during the fusion of 1kg of 11H to be 39.1495×1026MeV


  1. The Fission of 1.0kg of 235U in a fission reactor.

Ans: We are given:

Amount of 92U235=1000gm

1 mole, i.e., 235g of 92235U contains 6.023×1023atoms.

1000g of 92235U contains 6.023×1023×1000235atoms

We know that the amount of energy released in the fission of one atom of 92235U is200MeV. Therefore, energy released from the fission of 1kg of 92235U is:

E2=6×1023×1000×200235=5.106×1026MeV

E1E2=39.1495×10265.106×1026=7.678

Hence, we found the energy released during the fusion of 1kg of hydrogen is nearly 8 times the energy released during the fusion of 1kg of uranium. 


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: We are given the following:

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

Amount of nuclear power, P1=10100×2×105=2×104MW

P1=2×104×106J/s=2×1010×3600×24×365J/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:

25100×200=50MeV=50×1.6×1019×106=8×1012J

The number of atoms required for fission per year would be:

2×1010×60×60×24×3658×1012=78840×1024atoms

1 mole, i.e., 235g of U235contains 6.023×1023atoms

That is, the mass of 6.023×1023atomsof U235=235g=235×103kg

Also, the mass of:78840×1024atoms of U235=235×1036.023×1023×78840×1024=3.076×104kg

Hence, the mass of uranium needed per year is found to be,3.076×104kg.


NCERT Solutions for Class 12 Physics - Free PDF Download

To ace in your paper, you need to understand all the basic concepts of your subjects. For that, solutions need to be very simplified in a clear format with the help of comparatively easy methods and lesser calculation. This saves time and effort in the exam. There are live classes available for online mode of study and understanding concepts. Also, in offline mode simple free pdf are available too. Thus, learning and writing notes go simultaneously. You can have a complete series of notes by downloading these free PDFs.


Solving Chapter 13 for Class 12 Physics chapter Nuclei in this manner will bring you maximum marks. For this, Vedantu’s experienced faculty has prepared these solutions by themselves with a lot of attention. The solution is available in pdf format also. You can download and access NCERT Class 12 Physics Chapter 13 pdf offline, too.


So, just grab them before heading towards the main exams. The study material is present in an enormous variety. You can choose the one which is favourable for you and quickly adaptable.

This chapter mostly concerns atomic energy and its constituents. It includes a simple introduction about atomic nuclei, atomic masses, their composition, size of nucleus, mass number and binding energy, strong relationships, nuclear force, radioactivity, and nuclear energy.


There is a greater numerical portion in the chapter and lots of numbers that you have to remember. It deals with the foundation of nuclear energy. How binding energy and mass numbers play a crucial role in the generation of nuclear power. There are many questions in the chapter and if you want a simplified solution of these questions in an easily understandable way, then download the free pdf of Vedantu consisting of all Chapter 13.


NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei

Chapter 13 Nuclei

Most of the questions of Physics are in numerical format. Thus, practice and regular revision make a good command of the subject. The NCERT Books Soutions for Class 12 Physics Chapter 13 Nuclei are a moderate level, hence not too tough if you think. All you need to be a little bit conceptual, practising, and visiting a variety of questions by taking Chapter 13 as reference material.


Benefits of NCERT Solutions for Class 12th Physics

Chapter 13 of Class 12th Physics is not very tough and not very easy. After learning concepts and practising the maximum number of questions, you would find them of a moderate level to solve. Some benefits of Chapter 13 are listed below:


  • Chapter 13 reveals all possible methods of solving concerned problems.

  • Chapter 13 routine practice improves marks and performance.

  • These solutions are created by expertise; hence they are completely accurate.

  • The solutions cover almost all the tips and tricks for quick solving.

  • Chapter 13 makes you learn all the important formulas at one hand only.

  • These solutions are well explained in detail with logics; hence each solution explains itself the method of solving.


How Would Vedantu Study Material Help Students?

  • Vedantu's study material is designed to improve students' problem-solving ability.

  • The study material includes a range of easy, moderate, and tough question series to cater to different students' needs.

  • NCERT Solutions for Class 12 Physics Chapter 13 include both expected and previous year solved questions.

  • The study material helps students understand the distribution of marks for different types of questions.

  • The study material offers test series, quizzes, sample papers, and pattern-wise differentiated questions.

  • The study material is created by experienced and expert faculty members of Vedantu.

  • Chapter 13 study material is reliable, accurate, and simplified.

  • The solutions are aligned with the latest CBSE guidelines and syllabus, and the study material is regularly updated.

  • Using a single source of study material saves time and energy while enhancing problem-solving skills.

  • The last chapter covered in CBSE Class 12 Term II Physics syllabus is Nuclei.

  • NCERT Solutions for Chapter 13 help students solve most questions related to Nuclei.

  • The chapter covers topics such as composition and size of nucleus, mass-energy relationship, nuclear force, nuclear fusion, and nuclear fission.

  • Vedantu provides important questions and short key-notes for students preparing this chapter.

  • Students can download and refer to relevant study materials for Chapter 13 from Vedantu.


Conclusion

The NCERT Solutions for Class 12 Physics Chapter 13 Nuclei offered by Vedantu provide a comprehensive and effective resource for students. The study material is meticulously designed to enhance students' problem-solving abilities, catering to students of all levels. With a range of easy, moderate, and tough question series, along with previous year solved questions, students can gain a thorough understanding of the mark-wise distribution of questions. The study material, created by experienced faculty members, is reliable, accurate, and regularly updated to align with the latest CBSE guidelines and syllabus. By practicing from a single source, students can save time and energy while improving their capability to solve a plethora of questions. Vedantu's NCERT Solutions and additional study materials are highly beneficial, equipping students with the necessary knowledge and skills to excel in Chapter 13 Nuclei.


Other Study Material for CBSE Class 12 Physics Chapter 13



Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. You can use it as your 12th Physics guide. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Related Links for NCERT Class 12 Physics in Hindi

Discover relevant links for NCERT Class 12 Physics in Hindi, offering comprehensive study materials, solutions, and resources to enhance understanding and aid in exam preparation.




Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


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FAQs on NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

1. Which website provides NCERT Solutions for Class 12 Physics Chapter 13 Nuclei?

Vedantu provides NCERT Solutions for Class 12 Physics Chapter 13 Nuclei. It is available in free PDF format. NCERT Solutions for Chapter 13 Nuclei are designed by expert Physics tutors at Vedantu. The material is designed to help students revise the complete chapter and score well in exams. NCERT Solutions for Class 12 Physics Chapter 13 are prepared as per the latest NCERT norms and exam guidelines. It provides students with a thorough knowledge of the chapter and simple explanations to the exercise problems with an emphasis on critical concepts. It is designed to help students in doubt clearance.

2. What are the benefits of NCERT Solutions for Class 12 Physics Chapter 13 Nuclei?

CBSE Class 12 Physics Chapter 13 Nuclei is an important chapter from exam perspective. NCERT Solutions explain the concepts related to Nuclei and answer questions related to the chapter. It is important to understand the chapter thoroughly to score well in exams. NCERT Solutions for Class 12 Physics Chapter 13 are the most reliable online resource. Students facing any doubt in the chapter can seek guidance from expert solutions. The tutors who have prepared the solutions have years of experience in the field of teaching. NCERT Solutions for Class 12 Chapter 13 Nuclei is a great way to be a top scorer in the Physics exam.

3. What are the salient aspects of NCERT Solutions for Class 12 Physics Chapter 13 Nuclei by Vedantu?

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are available on its site. Following are some of the key features of NCERT Solutions for Class 12 Physics Chapter 13 offered by Vedantu:

  • As per the latest syllabus and the exam pattern.

  • Step-by-step solutions.

  • Prepared by subject matter experts.

  • Free PDF material.

  • Complete coverage of the NCERT questions.

4. Can Vedantu Help in Achieving Maximum Marks in Board Examinations?

Vedantu is present to bring the best out of you through the best faculty of each topic they have. The solutions are appropriately created according to the latest CBSE guidelines. So, yes after practising all the questions and downloading Nuclei Class 12 Chapter 13 pdf from Vedantu, you can achieve maximum marks.

5. What is the nucleus definition?

The nucleus is the centre of an atom. It consists of positively charged particles called protons and electrically neutral particles called neutrons. Protons and neutrons form the nucleus. The atomic number is the number of protons of the nucleus.

6. What does Chapter 13 of Class 12 Physics tell us about?

Chapter 13 of Class 12 Physics is "Nuclei". The chapter mainly focuses on atoms and their constituents. It includes an Introduction About Atoms, Nuclei, Atomic Masses, Size Of Nucleus, Mass Number, Binding Energy, Nuclear Force, Nuclear Energy And Radioactivity. The chapter also consists of many numerical and examples for a better understanding of the chapter. 

7. How to make a fruitful study plan for Chapter 13 of Class 12 Physics?

The given technique will help students in making a better study plan for Chapter 13 of Class 12 Physics:

  • Build a timetable so that you can give your full attention to Chapter 13 of Class 12 Physics.

  • Study the chapter for at least 2 hours a day.

  • Solve all the NCERT back questions and examples to understand the chapter clearly.

  • Prepare notes for learning important terminologies used in the chapter.

  • Try to avoid studying for longer periods.

  • For better concentration, meditate daily to keep your mind fit.

  • Also, make use of other reference books for solving a variety of questions.

  • You can utilize the Vedantu Mobile app to study anytime.

8. How to score well in Chapter 13 of Class 12 Physics?

A student of Class 12 should make notes of Chapter 13 of Class 12 Physics for better understanding. He or she should go through the chapter thoroughly. Practising numerical and examples can help students to have a strong grip on that chapter. Regular revision of a particular chapter is a must. Solving question papers can also be helpful to students. They can also take the help of Vedantu as it provides NCERT Solutions of Chapter 13 of Class 12 Physics plus the question papers, available for free. Following the above steps, a student can score good marks in Class 12 Physics Exams.

9. What method should be kept in mind while downloading the PDF file of Physics Chapter 13 Class 12? 

The steps are given below to download the PDF file of Physics Chapter 13 Class 12 are; 

  • Tap Chapter 13 of Class 12 Physics. The link will open.

  • The link will land you on Vedantu's website

  • At the header of the page, you will see the "Download PDF" option.

  • After clicking on that option the NCERT Solutions PDF file will get downloaded.

  • You can now solve the questions offline also.

In this PDF file, the solutions are written in very simple language, so that the students can understand them easily.

10. How can you express the terms "Binding Energy" and "Binding Energy per Nucleon"?

The word binding energy is described as the energy that is needed to shatter a nucleus in its constituent particles (neutrons and protons). The nucleus is broken in such a way that there is a huge separation between the constituent particles, so that they do not interact with each other. The average energy, which is used to take out one nucleon from the nucleus is known as binding energy per nucleon.