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**NCERT QUESTION**

**(Nuclei)****Ques 13.1:****(a) Two stable isotopes of lithium **** and **** have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.****(b) Boron has two stable isotopes, **** and ****. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of **** and ****.****Ans: (a)** Mass of lithium isotope , *m*_{1} = 6.01512 u

Mass of lithium isotope , *m*_{2} = 7.01600 u

Abundance of , *η*_{1}= 7.5%

Abundance of , *η*_{2}= 92.5%

The atomic mass of lithium atom is given as:**(b)** Mass of boron isotope , *m*_{1} = 10.01294 u

Mass of boron isotope , *m*_{2} = 11.00931 u

Abundance of , *η*_{1} = *x*%

Abundance of , *η*_{2}= (100 − *x*)%

Atomic mass of boron, *m* = 10.811 u

The atomic mass of boron atom is given as:

And 100 − *x* = 80.11%

Hence, the abundance of is 19.89% and that of is 80.11%.

**Ques 13.2:****The three stable isotopes of neon: ****and **** have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.****Ans: **Atomic mass of , *m*_{1}= 19.99 u

Abundance of , *η*_{1} = 90.51%

Atomic mass of , *m*_{2} = 20.99 u

Abundance of , *η*_{2} = 0.27%

Atomic mass of , *m*_{3} = 21.99 u

Abundance of , *η*_{3} = 9.22%

The average atomic mass of neon is given as:

= 20.1771 u

**Ques 13.3:****Obtain the binding energy (in MeV) of a nitrogen nucleus ****, given ****=14.00307 u****Ans: **Atomic mass of nitrogen , *m* = 14.00307 u

A nucleus of nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δ*m* = 7*m** _{H}* 7

Where,

Mass of a proton,

Mass of a neutron,

∴Δ

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c

∴Δ

Hence, the binding energy of the nucleus is given as:

Where,

∴

= 104.66334 MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

**Ques 13.4:****Obtain the binding energy of the nuclei **** and **** in units of MeV from the following data:**** = 55.934939 u , ****= 208.980388 u****Ans: **Atomic mass of , *m*_{1} = 55.934939 u

nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, Δ*m* = 26 × *m** _{H}* 30 ×

Where,

Mass of a proton,

Mass of a neutron,

∴Δ

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c

∴Δ

The binding energy of this nucleus is given as:

Where,

∴

= 492.26 MeV

Average binding energy per nucleon

Atomic mass of ,

nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δ

Where,

Mass of a proton,

Mass of a neutron,

∴Δ

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/

∴Δ

Hence, the binding energy of this nucleus is given as:

= 1.760877 × 931.5

= 1640.26 MeV

Average binding energy per nucleon =

**Ques 13.5:****A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of **** atoms (of mass 62.92960 u).****Ans: **Mass of a copper coin, *m*’ = 3 g

Atomic mass of atom, *m* = 62.92960 u

The total number of atoms in the coin

Where,

N_{A} = Avogadro’s number = 6.023 × 10^{23} atoms /g

Mass number = 63 g

nucleus has 29 protons and (63 − 29) 34 neutrons

∴ Mass defect of this nucleus, Δ*m*' = 29 × *m** _{H}* 34 ×

Where,

Mass of a proton,

Mass of a neutron,

∴ Δ

= 0.591935 u

Mass defect of all the atoms present in the coin, Δ

= 1.69766958 × 10

But 1 u = 931.5 MeV/

∴Δ

Hence, the binding energy of the nuclei of the coin is given as:

= 1.69766958 × 10

= 1.581 × 10

But 1 MeV = 1.6 × 10

= 2.5296 × 10

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

**Ques 13.6:****Write nuclear reaction equations for****(i) α-decay of **

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

**Ques 13.7:****A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?**

Original amount of the radioactive isotope =

It is given that only 3.125% of

Where,

λ = Decay constant

Hence, the isotope will take about 5*T* years to reduce to 3.125% of its original value.

**(b)** After decay, the amount of the radioactive isotope = *N*

It is given that only 1% of *N*_{0} remains after decay. Hence, we can write:

Since, *λ* = 0.693/*T*

Hence, the isotope will take about 6.645 *T* years to reduce to 1% of its original value.**Ques 13.8:****The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive **** present with the stable carbon isotope **** . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ****, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of **** dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization.****Ans: **Decay rate of living carbon-containing matter, *R* = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of , = 5730 years

The decay rate of the specimen obtained from the Mohenjodaro site:*R*' = 9 decays/min

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

ohenjodaro period.

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

Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.**Ques 13.9:****Obtain the amount of **** necessary to provide a radioactive source of 8.0 mCi strength. The half-life of **** is 5.3 years.****Ans: **The strength of the radioactive source is given as:

Where,*N* = Required number of atoms

Half-life of , = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10^{8} s

For decay constant *λ*, we have the rate of decay as:

Where, *λ*

For :

Mass of 6.023 × 10^{23} (Avogadro’s number) atoms = 60 g

∴Mass of atoms

Hence, the amount of necessary for the purpose is 7.106 × 10^{−6} g.

**Ques 13.10:****The half-life of **** is 28 years. What is the disintegration rate of 15 mg of this isotope?****Ans: **Half life of , = 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 10^{8} s

Mass of the isotope, *m* = 15 mg

90 g of atom contains 6.023 × 10^{23} (Avogadro’s number) atoms.

Therefore, 15 mg of contains:

Rate of disintegration,

Where,

λ = Decay constant

Hence, the disintegration rate of 15 mg of the given isotope is

7.878 × 10^{10} atoms/s.**Ques 13.11:****Obtain approximately the ratio of the nuclear radii of the gold isotope **** and the silver isotope ****.****Ans: **Nuclear radius of the gold isotope = *R*_{Au}

Nuclear radius of the silver isotope = *R*_{Ag}

Mass number of gold, *A*_{Au} = 197

Mass number of silver, *A*_{Ag} = 107

The ratio of the radii of the two nuclei is related with their mass numbers as:

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.**Ques 13.12:****Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) **

**Given **** = 226.02540 u, **** = 222.01750 u,****= 220.01137 u,****= 216.00189 u.****Ans: (a)** Alpha particle decay of emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

*Q*-value of emitted *α*-particle = (Sum of initial mass − Sum of final mass) *c*^{2}

Where,*c* = Speed of light

It is given that:

*Q*-value = [226.02540 − (222.01750 4.002603)] u *c*^{2}

= 0.005297 u *c*^{2}

But 1 u = 931.5 MeV/*c*^{2}

∴*Q* = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the *α*-particle

**(b)** Alpha particle decay of is shown by the following nuclear reaction.

It is given that:

Mass of = 220.01137 u

Mass of = 216.00189 u

∴*Q*-value =

≈ 641 MeV

Kinetic energy of the *α*-particle

= 6.29 MeV**Ques 13.13:****The radionuclide ^{11}C decays according to**

**calculate Q and compare it with the maximum energy of the positron emitted**

Atomic mass of = 11.011434 u

Atomic mass of

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the *Q*-value (Δ*Q*) of the nuclear masses of the nucleus is given as:

Where,*m** _{e}* = Mass of an electron or positron = 0.000548 u

If atomic masses are used instead of nuclear masses, then we have to add 6

Hence, equation (1) reduces to:

∴Δ

= (0.001033

But 1 u = 931.5 Mev/

∴Δ

The value of

emission of the nucleus is given as:

It is given that:

Atomic mass of = 22.994466 u

Atomic mass of = 22.989770 u

Mass of an electron,

There are 10 electrons in and 11 electrons in . Hence, the mass of the electron is cancelled in the

∴ Q = [ 22.994466 - 22.989770]c

= (0.004696 c

But 1 u = 931.5 Me V/c

∴ Q = 0.004696 x 931.5 = 4.374 Me V

The daughter nucleus is too heavy as compared to and . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the

It is given that:

Atomic mass

Atomic mass

Atomic mass

According to the question, the

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

It is given that:

Atomic mass of

Atomic mass of

Atomic mass of

The

The positive

It is given that:

Atomic mass of = 55.93494 u

Atomic mass of

The

The

**Ques 13.17:****The fission properties of **** are very similar to those of ****. ****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 **** undergo fission?****Ans: **Average energy released per fission of ,

Amount of pure , *m* = 1 kg = 1000 g

N_{A}= Avogadro number = 6.023 × 10^{23}

Mass number of = 239 g

1 mole of contains N_{A} atoms.

∴ *m* g of contains

∴Total energy released during the fission of 1 kg of is calculated as:-

Hence, is released if all the atoms in 1 kg of pure undergo fission.

**Ques 13.18:****A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much **** did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of **** and that this nuclide is consumed only by the fission process.****Ans: **Half life of the fuel of the fission reactor, years

= 5 × 365 × 24 × 60 × 60 s

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

1 mole, i.e., 235 g of contains 6.023 × 10^{23} atoms.

∴1 g contains

The total energy generated per gram of is calculated as:

The reactor operates only 80% of the time.

Hence, the amount of consumed in 5 years by the 1000 MW fission reactor is calculated as:

∴ Initial amount of = 2 × 1538 = 3076 kg

**Ques 13.19:****How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as****Ans: **The given fusion reaction is:

Amount of deuterium, *m* = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 10^{23} atoms.

∴2.0 kg of deuterium contains

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴ Total energy per nucleus released in the fusion reaction:

Power of the electric lamp, *P* = 100 W = 100 J/s.

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

**Ques 13.20:****Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)****Ans: **When two deuterons collide head-on, the distance between their centres, *d* is given as:

Radius of 1^{st} deuteron Radius of 2^{nd} deuteron

Radius of a deuteron nucleus = 2 fm = 2 × 10^{−15} m

∴ *d* = 2 × 10^{−15} 2 × 10^{−15} = 4 × 10^{−15} m

Charge on a deuteron nucleus = Charge on an electron = *e* = 1.6 × 10^{−19} C

Potential energy of the two-deuteron system:

Where,

= Permittivity of free space

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.**Ques 13.21:****From the relation R = R_{0}A^{1}/^{3}, where R_{0} 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).**

Where,

Nuclear matter density,

Let

Hence, mass of the nucleus =

Hence, the nuclear matter density is independent of

**Show that if **** 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 *Q*_{1}. The nuclear reaction can be written as:

Let the amount of energy released during the positron capture process be *Q*_{2}. The nuclear reaction can be written as:

= Nuclear mass of

= Nuclear mass of

= Atomic mass of

= Atomic mass of *m** _{e}* = Mass of an electron

It can be inferred that if

In other words, this means that if emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the

Mass of magnesium isotope ,

Mass of magnesium isotope ,

Mass of magnesium isotope ,

Abundance of ,

Abundance of ,

Hence, abundance of ,

We have the relation for the average atomic mass as:

2431,2 = 1894.5783096 + 24.98584 x + 545.8942159 - 25.98259 x

0.99675* = 9.2725255

∴ x ≈ 9.3%

And 21.01 -* = 11.71%

Hence, the abundance of is 9.3% and that of is 11.71%.

**Ques 13.24:****The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei **** and **** from the following data:**

=** 39.962591 u**** = 40.962278 u****= 25.986895 u**** = 26.981541 u****Ans: **For

For

A neutron is removed from a nucleus. The corresponding nuclear reaction can be written as:

It is given that:

Mass = 39.962591 u

Mass = 40.962278 u

Mass = 1.008665 u

The mass defect of this reaction is given as:

Δ*m* =

∴Δ*m* = 0.008978 × 931.5 MeV/*c*^{2}

Hence, the energy required for neutron removal is calculated as:

For , the neutron removal reaction can be written as:

It is given that:

Mass = 26.981541 u

Mass = 25.986895 u

The mass defect of this reaction is given as:

Hence, the energy required for neutron removal is calculated as:

**Ques 13.25:****A source contains two phosphorous radio nuclides ****( T_{1/2 =} 14.3d) and **

Half life of ,

nucleus decay is 10% of the total amount of decay.

The source has initially 10% of nucleus and 90% of nucleus.

Suppose after

**Initially:**

Number of nucleus = *N*

Number of nucleus = 9 *N***Finally:**

Number of

Number of

For nucleus, we can write the number ratio as:

For , we can write the number ratio as:

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

Hence, it will take about 208.5 days for 90% decay of .**Ques 13.26:****Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:**

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

We know that:

Mass of

Mass of

Mass of ,

Hence, the

= (223.01850 − 208.98107 − 14.00324)

= (0.03419

But 1 u = 931.5 MeV/

∴

= 31.848 MeV

Hence, the

Now take a emission nuclear reaction:

We know that:

Mass of

Mass of

Mass of ,

= (223.01850 − 219.00948 − 4.00260) C

= (0.00642

= 0.00642 × 931.5 = 5.98 MeV

Hence, the *Q* value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.**Ques 13.27:****Consider the fission of **** 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 **** and ****. Calculate Q for this fission process. The relevant atomic and particle masses are***m*** =238.05079 u***m*** =139.90543 u***m*** = 98.90594 u**

**Ans: **In the fission of , 10 *β*^{−} particles decay from the parent nucleus. The nuclear reaction can be written as:

It is given that:

Mass of a nucleus *m*_{1} = 238.05079 u

Mass of a nucleus *m*_{2} = 139.90543 u

Mass of a nucleus , *m*_{3} = 98.90594 u

Mass of a neutron *m*_{4} = 1.008665 u*Q*-value of the above equation,

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

= *m*_{1} − 92*m*_{e}

= *m*_{2} − 58*m*_{e}

= *m*_{3} − 44*m*_{e}

= *m*_{4}

But 1 u = 931.5 MeV / c^{2}

∴ O = 0.247995 x 931.5 = 23 1.007 MeV

Hence, the *Q*-value of the fission process is 231.007 MeV.**Ques 13.28:****Consider the D−T reaction (deuterium−tritium fusion)**** ****(a) Calculate the energy released in MeV in this reaction from the data:****= 2.014102 u****= 3.016049 u****(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3 kT/2); k = Boltzmann’s constant, T = absolute temperature.)**

It is given that:

Mass of ,

Mass of ,

Mass of

Mass of ,

= [2.014102 + 3.016049 − 4.002603 − 1.008665]

= [0.018883

But 1 u = 931.5 MeV/

∴

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

Charge on the deuterium nucleus =

Charge on the tritium nucleus =

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

Where,

∈

Hence, 5.76 × 10

However, it is given that:

KE

Where,

T = Temperature required for triggering the reaction

Hence, the gas must be heated to a temperature of 1.39 × 10

**Ans: **It can be observed from the given *γ*-decay diagram that *γ*_{1} decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to *γ*_{1}-decay is given as:*E*_{1} = 1.088 − 0 = 1.088 MeV*hν*_{1}= 1.088 × 1.6 × 10^{−19} × 10^{6} J

Where,*h* = Planck’s constant = 6.6 × 10^{−34} Js

ν_{1} = Frequency of radiation radiated by *γ*_{1}-decay

It can be observed from the given *γ*-decay diagram that *γ*_{2} decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to *γ*_{2}-decay is given as:*E*_{2} = 0.412 − 0 = 0.412 MeV*hν*_{2}= 0.412 × 1.6 × 10^{−19} × 10^{6} J

Where,

ν_{2} = Frequency of radiation radiated by *γ*_{2}-decay

It can be observed from the given *γ*-decay diagram that *γ*_{3} decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to *γ*_{3}-decay is given as:*E*_{3} = 1.088 − 0.412 = 0.676 MeV*hν*_{3}= 0.676 × 10^{−19} × 10^{6} J

Where,

ν_{3} = Frequency of radiation radiated by *γ*_{3}-decay

Mass of = 197.968233 u

Mass of = 197.966760 u

1 u = 931.5 MeV/*c*^{2}

Energy of the highest level is given as:

β_{1} decays from the 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the *β*_{1} particle = 1.3720995 − 1.088

= 0.2840995 MeV

β_{2} decays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the *β*_{2} particle = 1.3720995 − 0.412

= 0.9600995 MeV

**Ques 13.30:****Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ^{235}U in a fission reactor.**

1 mole, i.e., 1 g of hydrogen ( ) contains 6.023 × 10

∴1000 g of contains 6.023 × 10

Within the sun, four nuclei combine and form one nucleus. In this process 26 MeV of energy is released.

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

1 mole, i.e., 235 g of contains 6.023 × 10

∴1000 g of contains

It is known that the amount of energy released in the fission of one atom of is 200 MeV.

Hence, energy released from the fission of 1 kg of is:

∴

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium.**Ques 13.31: ****Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ^{235}U to be about 200MeV.**

10% of this amount has to be obtained from nuclear power plants.

∴ Amount of nuclear power,

= 2 × 10

= 2 × 10

= 2 × 10

Heat energy released per fission of a

Efficiency of a reactor = 25%

Hence, the amount of energy converted into the electrical energy per fission is calculated as:

Number of atoms required for fission per year:

1 mole, i.e., 235 g of U^{235} contains 6.023 × 10^{23} atoms.

∴ Mass of 6.023 × 10^{23} atoms of U^{235} = 235 g = 235 × 10^{−3} kg

∴ Mass of 78840 × 10^{24} atoms of U^{235}

Hence, the mass of uranium needed per year is 3.076 × 10^{4} kg.

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