Ncert Solution for Class 12 Physics Chapter 13 Nuclei

Nuclei ncert solutions: Ncert Class 12 Physics Chapter 13 Exercise Solutions

Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 13
Chapter Name	Nuclei class 12 exercise solutions
Category	Ncert Solutions
Medium	English

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Question 13.1: Obtain the binding energy (in MeV) of a nitrogen nucleus ( \(_{7}^{14}{N}\) ) , given m ( \(_{7}^{14}{N}\) ) = 14.00307 u.

Solution 13.1:

Atomic mass of nitrogen \(_{7}^{14}{N}\) , m = 14.00307 u

A nucleus of \(_{7}^{14}{N}\) nitrogen contains 7 neutrons and 7 protons.

∆m = 7m_H + 7m_n − m is the mass defect of the nucleus

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But, 1 u = 931.5 MeV/c²

∆m = 0.11236 × 931.5 MeV/c²

E_b = ∆mc² is the binding energy of the nucleus

Where, c = Speed of light

\(E_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)

= 104.66334 Mev

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

Question 13.2: Obtain the binding energy of the nuclei \(_{26}^{56}{Fe}\) and \(_{83}^{209}{Bi}\) in units of MeV from the following data:
(a) m ( \(_{23}^{56}{Fe}\) ) = 55.934939 u
(b) m( \(_{83}^{209}{Bi}\) ) = 208.980388 u

Solution 13.2:

(a) Atomic mass of \(_{26}^{56}{Fe}\) , m₁ = 55.934939 u

\(_{26}^{56}{Fe}\) nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = 26 × m_H + 30 × m_n − m₁

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But, 1 u = 931.5 MeV/c²

∆m = 0.528461 × 931.5 MeV/c²

\(E_{b_{1}}\) = ∆mc² is the binding energy of the nucleus.

Where, c = Speed of light

\(E_{b_{1}} = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)

= 492.26 MeV

Therefore, the average binding energy per nucleon =

\(\frac{492.26}{56} = 8.79 MeV\)

(b) Atomic mass of \(_{83}^{209}{Bi}\) , m₂ = 208.980388 u

\(_{83}^{209}{Bi}\) nucleus has 83 protons and (209 − 83) 126 neutrons.

The mass defect of the nucleus is given as:

∆m’ = 83 × m_H + 126 × m_n − m₂

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But, 1 u = 931.5 MeV/c²

∆m’ = 1.760877 × 931.5 MeV/c²

E_b2 = ∆m’c² is the binding energy of the nucleus.

= 1.760877 × 931.5

\( \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\)

= 1640.26 MeV

Therefore, the average binding energy per nucleon = 1640.26/209 = 7.848 MeV.

Question 13.3: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of
\(_{29}^{63}{Cu}\) atoms (of mass 62.92960 u).

Solution 13.3:

Mass of a copper coin, m’ = 3 g

Atomic mass of \(_{29}^{63}{Cu}\) atom, m = 62.92960 u

The total number of \(_{29}^{63}{Cu}\) atoms in the coin, \(N = \frac{N_A \times m’}{Mass \; number}\)

Where,

N_A = Avogadro’s number = 6.023 × 10²³ atoms /g

Mass number = 63 g

\(N = \frac{6.023 \times 10^{23}\times 3}{63}\) = 2.868 x 10²² atoms

\(_{29}^{63}{Cu}\) nucleus has 29 protons and (63 − 29) 34 neutrons

Mass defect of this nucleus, ∆m’ = 29 × m_H + 34 × m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 10²²

= 1.69766958 × 10²² u

But, 1 u = 931.5 MeV/c²

∆m = 1.69766958 × 10²² × 931.5 MeV/c²

E_b = ∆mc² is the binding energy of the nuclei of the coin

= 1.69766958 × 10²² × 931.5 MeV/c² × c²

= 1.581 × 10²⁵ MeV

But 1 MeV = 1.6 × 10⁻¹³ J

E_b = 1.581 × 10²⁵ × 1.6 × 10⁻¹³

= 2.5296 × 10¹² J

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

Question 13.4: Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197 }{Au}\) and the \(_{ 47 }^{ 107 }{Ag}\) silver isotope.

Solution 13.4:

Nuclear radius of the gold isotope \(_{ 79 }^{ 197 }{Au}\) = R_Au

Nuclear radius of the silver isotope \(_{ 47 }^{ 107 }{Ag}\) = R_Ag

The mass number of gold, A_Au = 197

The mass number of silver, A_Ag = 107

Following is the relationship between the radii of the two nuclei and their mass number:

\(\frac{ R_{ Au }}{ R_{ Ag }}=\left (\frac{ R_{ Au }}{ R_{ Ag }} \right )^{\frac{ 1 }{ 3 } }\)

\(=\left (\frac{ 197 }{ 107 } \right )^{\frac{ 1 }{ 3 } }\)

= 1.2256

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

Question 13.5: The Q value of a nuclear reaction A + b → C + d is defined by Q = [ m_A+ m_b− m_C− m_d]c^2,  where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) \(_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}\)
(ii) \(_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}\)
Atomic masses are given to be
\(m( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u\)
\(m( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u\)
\(m( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u\)
\(m( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u\)

Solution 13.5:

(i) The given nuclear reaction is:

\(_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}\)

It is given that:

Atomic mass \(m( _{ 1 }^{ 1 }{H} ) = \; 1.007825 \; u\)

Atomic mass \(m( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u\)

Atomic mass \(m( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u\)

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

Q = [ \(m( _{ 1 }^{ 1 }{H} ) \) + \(m( _{ 1 }^{ 3 }{H} ) \) – 2m\(m( _{ 1 }^{ 2 }{H} )\)] c²

= [ 1.007825 + 3.016049 – 2 x 2.014102] c²

Q = ( – 0.00433 c²) u

But 1 u = 931.5 MeV/c²

Q = -0.00433 x 931.5 = – 4.0334 MeV

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

(ii) The given nuclear reaction is:

\(_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}\)

It is given that:

Atomic mass of \(m( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u\)

Atomic mass of m \(m( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u\)

Atomic mass of \(m( _{ 2 }^{ 4 }{He} ) = \; 4.002603 \; u\)

The Q-value of this reaction is given as:

Q = [ \(2m( _{ 6 }^{ 12 }{C} ) \) – \(m( _{ 10 }^{ 20 }{Ne} )\) – \(m( _{ 2 }^{ 4 }{He} )\) ] c²

= [ 2 x 12.000000 – 19.992439 – 4.002603 ] c²

= [ 0.004958 c²] u

= 0.004958 x 931.5 = 4.618377 MeV

Since we obtained a positive Q-value, it can be concluded that the reaction is exothermic.

Question 13.6: Suppose, we think of fission of a \(_{ 26 }^{ 56 }{Fe}\) nucleus into two equal fragments, \(_{ 13 }^{ 28 }{Al}\) . Is fission energetically possible? Argue by working out the Q of the process. Given,\(m( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u\) and \(m( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u\).

Solution 13.6:

The fission of \(_{ 26 }^{ 56 }{Fe}\) can be given as :

\(_{ 26 }^{ 56 }{Fe}\) -> 2\(_{ 13 }^{ 28 }{Al}\)

It is given that:

Atomic mass of \(m( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u\)

Atomic mass of \(m( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u\)

The Q-value of this nuclear reaction is given as:

Q = [ \(m( _{ 26 }^{ 56 }{Fe} ) \) – 2m\(( _{ 13 }^{ 28 }{Al} ) \) ] c²

= [ 55.93494 – 2 x 27.98191 ]c²

= ( -0.02888 c²) u

But, 1 u = 931.5 MeV/c²

Q = – 0.02888 x 931.5 = -26.902 MeV

Since the Q-value is negative for fission, it is energetically not possible.

Question 13.7: The fission properties of \(_{ 94 }^{ 239 }{ Pu }\) of are very similar to those of \(_{ 92 }^{ 235 }{ U }\) . The average energy released per fission is 180 MeV. How much energy is released if all the atoms in 1 kg of pure \(_{ 94 }^{ 239 }{ Pu }\) undergo fission?

Solution 13.7:

Average energy released per fission of \(_{ 94 }^{ 239 }{ Pu }\) , E_av  = 180 MeV

Amount of pure \(_{ 94 }^{ 239 }{ Pu }\), m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of \(_{ 94 }^{ 239 }{ Pu }\) = 239 g

1 mole of \(_{ 94 }^{ 239 }{ Pu }\) contains N_A atoms.

Therefore, mg of \(_{ 94 }^{ 239 }{ Pu }\) contains \(\left ( \frac{N_A}{Mass \; Number}\times m \right )\) atoms

\(\frac{6.023 \times 10^{23}}{239}\times 1000 \)

\(= 2.52 \times 10^{ 24 }\) atoms

The total energy released during the fission of 1 kg of \(_{ 94 }^{ 239 }{ Pu }\) is calculated as:

E = E_av x 2.52 x 10²⁴

= 180 x 2.52 x 10²⁴

= 4.536 x 10²⁶ MeV

Hence, 4.536 x 10²⁶ MeV is released if all the atoms in 1 kg of pure \(_{ 94 }^{ 239 }{ Pu }\) undergo fission.

Question 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

Solution 13.8:

The fusion reaction given is

Amount of deuterium, m = 2 kg

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

Therefore, 2 Kg of deuterium contains (6.023 x 10²³/2) x 2000 = 6.023 x 10²⁶ atoms

From the reaction given, it can be understood that 2 atoms of deuterium fuse, 3.27 MeV energy are released. The total energy per nucleus released during the fusion reaction is

E = (3.27/2) x 6.023 x 10²⁶  MeV

= (3.27/2) x 6.023 x 10²⁶  x 1.6 x 10 ^-19 x 10⁶

= 1.576 x 10¹⁴ J

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

The energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is 1.576 x 10¹⁴ /100 = 1.576 x 10¹²s

= (1.576 x 10¹²)/ (365 x 24 x 60 x 60)

= (1.576 x 10¹²)/3.1536 x10⁷

= 4.99 x 10⁴ years.

Question 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.)

Solution 13.9:

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

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2 f m = 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

The potential energy of the two-deuteron system: \(v = \frac{ e^{ 2 } }{ 4 \pi \epsilon_0 d }\)

Where,

ϵ_o = Permittivity of free space

\(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} Nm^2C^{-2}\)

Therefore,

\(V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} } J\)

\(V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} \times (1.6 \times 10^{-19})}eV\) = 360 k eV

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

Question 13.10: From the relation R = R₀A^1/3, 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).

Solution 13.10:

The nuclear radius is given as

R = R₀A^1/3

Here,

R₀ is Constant

A is the mass number of the nucleus

Nuclear matter density, ρ = Mass of the nucleus/Volume of the nucleus

Mass of the nucleus = mA

Density of the nucleus = (4/3)πR³

= (4/3)π(R₀A^1/3)³

= (4/3)πR₀³A

ρ = mA/[(4/3)πR₀³A]

= 3mA/(4πR₀³A)

ρ = 3m/(4πR₀³)

Hence, nuclear matter density is nearly a constant and is independent of A.