Class 12 - Physics - Nuclei

You may find the following data useful in solving the exercises:

e = 1.6×10^{–19}C, N = 6.023×10^{23} per mole, 1/ (4πε_{0}) = 9 × 10^{9}N m^{2}/C^{2}

k = 1.381×10^{–23}J^{0}K^{–1}, 1 MeV = 1.6×10^{–13}J 1 u = 931.5 MeV/c^{2}

1 year = 3.154×10^{7}s, m_{H} = 1.007825 u mn = 1.008665 u

m (^{4}_{2}He ) = 4.002603 u, m_{e} = 0.000548 u

**Question1.**

(a) Two stable isotopes of lithium (^{6}_{3}Li) and (^{7}_{3}Li) 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, (^{10}_{5}B) and (^{11}_{5}B). 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 ^{10}_{5}B and ^{11}_{5}B.

Answer:

Given:

Mass of (^{6}_{3}Li) lithium isotope, m_{1} = 6.01512 u

Mass of (^{7}_{3}Li) lithium isotope, m = 7.01600 u

Abundance of (^{6}_{3}Li), η_{1} =7.5%

Abundance of (^{7}_{3}Li), η_{2}= 92.5%

The atomic mass of lithium atom is given as:

m = (m_{1} η_{1} + m η_{2})/ (η_{1} + η_{2})

= (6.01512 x 7.5 + 7.01600 x92.5)/ (7.5 + 92.5)

=6.940934 u

(b) Mass of boron isotope (^{10}_{5}B), m = 10.01294 u

Mass of boron isotope (^{11}_{5}B), m = 11.00931 u

Abundance of (^{10}_{5}B), η_{1} = x%

Abundance of (^{11}_{5}B), η_{2}= (100 − x) %

Atomic mass of boron, m = 10.811 u.

The atomic mass of boron atom is given as:

m = (m_{1} η_{1} + m η_{2})/ (η_{1} + η_{2})

10.811 = (10.01294 x x + 11.00931(100-x))/(x + 100-x)

108.11 = (10.01294x + 1100.931 – 11.00931x)

Therefore, x = (19.821)/ (0.99637)

=19.89%

And 100 − x = 80.11%

Hence, the abundance of (^{10}_{5}B) is 19.89% and that of (^{11}_{5}B)is 80.11%.

**Question2.**

The three stable isotopes of neon: ^{20}_{10}Ne, ^{21}_{10}Ne and ^{22}_{10}Ne 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.

Answer:

Given:

Atomic mass of _{10}^{20}Ne, m_{1}= 19.99 u

Abundance of _{10}^{20}Ne, η_{1} = 90.51%

Atomic mass of _{10}^{21}Ne, m_{2} = 20.99 u

Abundance of _{10}^{21}Ne, η_{2} = 0.27%

Atomic mass of _{10}^{22}Ne, m_{3} = 21.99 u

Abundance of _{10}^{22}Ne, η_{3} = 9.22%

The average atomic mass of neon is given as:

m= (m_{1} η_{1}+ m_{2} η_{2}+ m_{3} η_{3})/ (η_{1}+ η_{2}+ η_{3})

= (19.99x90.51+20.99x0.27+21.99x9.22)/ (90.51+0.27+9.22)

=20.1771u

** Question 3.**

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

Answer:

Given:

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

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

Hence, the mass defect of this nucleus, ∆m = 7m_{H} + 7m_{n} − m

Where,

Mass of a proton, m_{H} = 1.007825 u

Mass of a neutron, m_{n}= 1.008665 u

Therefore, ∆m = ((7 × 1.007825) + (7 × 1.008665) − 14.00307))

= (7.054775 + 7.06055 − 14.00307)

= 0.11236 u

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

∆m = (0.11236 × 931.5) MeV/c^{2}

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

E_{b} = ∆mc^{2}

Where, c = Speed of light

E_{b} = 0.11236 × 931.5(MeV/c^{2})/c^{2}

= 104.66334 MeV

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

**Question 4.**

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

Answer:

Given:

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

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

Hence, the mass defect of the nucleus, ∆m = ((26 × m_{H}) + (30 × m_{n}) – m_{Fe})

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/c^{2}

Therefore, ∆m = (0.528461 × 931.5) MeV/c^{2}

The binding energy of this nucleus is given as:

E_{b1} = ∆mc^{2}

Where, c =Speed of light

E_{b1} = (0.528461 × 931.5(MeV/c^{2})/c^{2})

= 492.26 MeV

Average binding energy per nucleon= (492.26/56)=8.76 MeV

Atomic mass of (^{209}_{83}Bi), m_{2} = 208.980388 u

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

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

∆m' = (83 × m_{H} + 126 × m_{n}) − m_{2}

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/c^{2}

Therefore, ∆m' = (1.760877 × 931.5) (MeV/c^{2}) x c^{2}

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

E_{b2} = ∆m'c^{2}

= (1.760877 × 931.5(MeV/c^{2}) x c^{2})

= 1640.26 MeV

Average binding energy per nucleon = (1640.26/209) =7.848MeV

**Question 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 ^{63}_{29}Cu atoms (of mass 62.92960 u).

Answer:

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 = (N_{A} x m’)/ (Mass number)

Where,

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

Mass number = 63 g

Therefore N = (6.023 × 10^{23} x3)/ (63) =2.868x10^{22} atoms.

(_{29}^{63}Cu) nucleus has 29 protons and (63-29) = 34 neutrons.

Therefore, 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

Therefore, ∆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 × 10^{22})

=1.69766958x10^{22}u.

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

Therefore, ∆m = (1.69766958 × 10^{22} × 931.5 (MeV/c^{2}))

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

E_{b}= ∆mc^{2}

= (1.69766958 × 10^{22} × 931.5(MeV/c^{2})/c^{2})

= 1.581 × 10^{25} MeV

But 1 MeV = 1.6 × 10^{−13} J

E_{b} = (1.581 × 10^{25} × 1.6 × 10^{−13})

= 2.5296 × 10^{12} J

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

**Question 6.**

Write nuclear reaction equations for

(i) α-decay of ^{226}_{88}Ra (ii) α-decay of ^{24}_{294}Pu

(iii) β^{–}-decay of ^{32}_{15}P (iv) β^{–}-decay of ^{210}_{83}Bi

(v) β^{+}-decay of ^{11}_{6}C (vi) β^{+}-decay of ^{97}_{43}Tc

(vii) Electron capture of ^{120}_{54}Xe

Answer:

α is a nucleus of helium and β is an electron (e^{−} for β^{−} and e^{+} for β^{+}). In every α decay, there is a loss of 2 protons and 4 neutrons.

In every β^{+}-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β^{−}-decay,

there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

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

- (
_{88}Ra^{226}) à (_{86}Rn^{222}) + (_{2}He^{4}) - (
_{94}Pu^{242}) à (_{92}U^{238}) +(_{2}He^{4}) - (
_{15}P^{32}) à (_{16}S^{32}) + e^{− }+ ̅ν - (
_{83}B^{210}) à (_{84}PO^{210}) + e^{-}+ ̅ν - (
_{6}C^{11}) à (_{5}B^{11}) + e^{+}+ ν - (
_{43}Tc^{97}) à (_{42}MO^{97}) + e^{+}+ ν - (
_{54}Xe^{120}) + e^{+}à (_{53}I^{120}) + ν

**Question 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?

Answer:

Given:

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N_{0}

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

It is given that only 3.125% of N_{0} remains after decay. Hence, we can write:

(N/N_{0})= 3.125 %=( 3.125)/ (100) = (1/32)

But (N/N_{0}) =e^{-}^{ λt}

Where, λ = Decay, constant t = Time

Therefore e^{- (}^{λt)} = (1/32)

- (λt) = ln (1) – ln (32)

- (λt) = 0 -3.4657

t= (3.4657)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (3.466)/ ((0.693)/ (T)) ≈ 5T years.

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

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

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

(N/N_{0})=1 %=( 1/100)

But (N/N_{0}) =e^{-}^{ λt}

Therefore e^{- (}^{λt)} = (1/100)

- (λt) = ln (1) – ln (100)

- (λt) = 0 - 4.6052

t= (4.6052)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (4.6052)/ ((0.693)/ (T)) =6.645T years.

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value

**Question 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 (^{14}_{6}C) present with the stable carbon isotope ^{12}_{6}C.

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 (5730years) of ^{14}_{6}C, and the measured activity, the age of the specimen can be approximately estimated.

This is the principle of ^{14}_{6}C dating used in archaeology.

Suppose a specimen from Mohenjo-Daro gives an activity of 9 decays per minute per gram of carbon.

Estimate the approximate age of the Indus-Valley civilisation.

Answer:

Given:

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

Half-life of (^{14}_{6}C), T_{ (1/2)} = 5730 years

The decay rate of the specimen obtained from the Mohenjo-Daro site:

R' = 9 decays/min

Let N' be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

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

(N/N’) = (R/R’) = e^{ (-λt)}

e^{ (-λt)}= (9/15) = (3/5)

(-λt) = log_{e} (3/5) = -0.5108

Therefore, t = (0.5108)/ (0.0.693)

But λ = (0.693)/ (T_{ (1/2)}) = (0.693)/ (5730)

Therefore, t = (0.5108)/ ((0.693/5730))

=4223.5 years

Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.

**Question 9.**

Obtain the amount of ^{60}_{27}Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ^{60}_{27}Co is 5.3 years.

Answer:

Given:

The strength of the radioactive source is given as:

(dN/dt) =8.0 mCi

= (8x10^{-3 }x 3.7x10^{10})

= (29.6x10^{7}) decay/s.

Where,

N = Required number of atoms

Half-life of (^{60}_{27}Co), T_{1/2} = 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:

(dN/dt) = (λN)

Where λ = (0.693/T_{1/2}) = (0.693)/ (1.67x10^{8}) s^{-1}

Therefore N = (1/ λ) (dN/dt)

= (29.6x10^{7})/ (0.693)/ (1.67x10^{8}) =7.133 x10^{16} atoms

For (^{60}_{27}Co):

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

Therefore, Mass of (7.133 x 10^{16}) atoms= (60x7.1333x10^{16})/ (6.023x10^{23})

=7.106x10^{-6}g

Hence, the amount of (^{60}_{27}Co) necessary for the purpose is 7.106 × 10^{−6} g.

**Question 10.**

The half-life of (^{90}_{38}Sr) is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer:

Half-life of (^{90}_{38}Sr), t_{ (1/2)} = 28 years

= (28 × 365 × 24 × 60 × 60)

= (8.83 × 10^{8}) s

Mass of the isotope, m = 15 mg

90 g of (^{90}_{38}Sr) atom contains 6.023 × 10^{23} (Avogadro’s number) atoms.

Therefore 15mg, of (^{90}_{38}Sr) contains:

= (6.023 x 10^{23} x 15 x 10^{-3})/ (90) i.e. 1.0038 x 10^{20} number of atoms.

Rate of disintegration, (dN/dt) = λ N

Where,

λ = Decay constant = (0.693)/ (8.83 x 10^{8}) s^{-1}

=7.878 x 10^{10 }atoms/s

Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 10^{10} atoms/s.

** Question 11.**

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

Answer:

Given:

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

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

(R_{Au}/ R_{Ag}) = (R_{Au}/ R_{Ag})^{ 1/3}

= (197/107)^{1/3}

=1.2256

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

**Question 12.**

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) ^{226}_{88} Ra and (b) ^{220}_{86} Rn.

Given m (22688 Ra) = 226.02540 u, m (^{222}_{86} Rn) = 222.01750 u,

m ( ^{222}_{86 }Rn ) = 220.01137 u, m ( ^{216}_{84} Po ) = 216.00189 u.

Answer:

(a) Alpha particle decay of (^{226}_{88}Ra) 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.

(^{226}_{88}Ra) à (^{222}_{88}Ra) + (^{4}_{2}He)

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:

m (^{226}_{88}Ra) =226.0250 u, m (^{222}_{86}Rn) =222.01750u, m (^{4}_{2}He) =4.002603u

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

= 0.005297 u c^{2}

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

Therefore Q = (0.005297 × 931.5) ≈ 4.94 MeV

Kinetic energy of the α-particle =

((Mass number after decay)/ (Mass number before decay) x Q)

= ((222)/ (226) x 4.94) = 4.85MeV.

(b) Alpha particle decay of (^{220}_{86}Rn)

(^{220}_{86}Rn) à (^{216}_{84}Po) + (^{4}_{2}He)

It is given that:

Mass of (^{220}_{86}Rn) = 220.01137 u

Mass of (^{216}_{84}Po) = 216.00189 u

Therefore, Q-value = [220.01137 – (216.00189+4.002603)] x931.5

≈ 641 MeV

Kinetic energy of the α-particle = ((220-4)/ (220) x 6.41)

= 6.29 MeV

**Question 13.**

The radionuclide ^{11}C decays according to

^{11}_{11}C à ^{11}_{5}B+ e^{+} + ν; T_{1/2} =20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values: m (^{11}_{6}C) = 11.011434 u and m (^{11}_{6}B) = 11.009305 u,

Calculate Q and compare it with the maximum energy of the positron emitted.

Answer:

The given nuclear reaction is:

^{11}_{11}C à ^{11}_{5}B+ e^{+} + ν

Atomic mass of, m (^{11}_{6}C) = 11.011434 u

Atomic mass of, m (^{11}_{6}B) = 11.009305 u

Half-life of ((^{11}_{6}C) nuclei, T (_{1/2)} = 20.3 min

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (∆Q) of the nuclear masses of the (^{11}_{6}C)

∆Q = [m (^{11}_{6}C) – [m (^{11}_{6}B) + m_{e}]] c^{2} (1)

Where, m_{e} = Mass of an electron or positron = 0.000548 u

c = Speed of light

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of C^{11 }and 5m_{e} in the case of ^{11}B.

Hence, equation (1) reduces to:

∆Q = [m (^{11}_{6}C) – m (^{11}_{6}B) -2 m_{e}]] c^{2}

Here, m (^{11}_{6}C) and m (^{11}_{6}B) are the atomic masses.

∆Q = [11.011434 − 11.009305 – (2 × 0.000548)] c^{2}

= (0.001033 c^{2}) u

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

∆Q = (0.001033 × 931.5)

≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.

**Question 14**.

The nucleus ^{23}_{10}Ne decays by β^{–}emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted.

Given that: m (^{23}_{10}Ne) = 22.994466 um (^{23}_{11}Na) = 22.089770 u.

Answer:

In β^{-} emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

β^{-} Emission of the nucleus (^{23}_{10}Ne)

(^{23}_{10}Ne) à (^{23}_{11}Na) + e^{-} + ̅ν + Q

It is given that:

Atomic mass m (^{23}_{10}Ne) of = 22.994466 u

Atomic mass m (^{23}_{11}Na) of = 22.989770 u

Mass of an electron, m_{e} = 0.000548 u

Q-value of the given reaction is given as:

Q= [m (^{23}_{10}Ne) + [m (^{23}_{11}Na) +m_{e}]] c^{2}

There are 10 electrons in and 11 electrons in (^{23}_{11}Na). Hence, the mass of the electron is cancelled in the Q-value equation.

Therefore Q = [22.994466 - 22.989770] c^{2}

= (0.004696 c^{2}) u.

But 1u = 9.31 (MeV/c^{2})

Therefore Q= (0.004696 x 931.5) =4.374MeV.

The daughter nucleus is too heavy as compared to e^{- }and ̅ν. Hence, it carries negligible energy.

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

** Question 15.**

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 + ^{3}_{1}H → ^{2}_{1}H + ^{2}_{1}H

(ii) ^{12}_{6}C + ^{11}_{6}C → ^{20}_{10}Ne + ^{4}_{2}He

Atomic masses are given to be

m (^{2}_{1}H) = 2.014102 u , m (^{3}_{1}H ) = 3.016049 u, m (^{12}_{6}C ) = 12.000000 u

m (^{20}_{10}Ne) = 19.992439 u

Answer:

Given:

- The given nuclear reaction is:

^{1}_{1}H + ^{3}_{1}H → ^{2}_{1}H + ^{2}_{1}H

It is given that:

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

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

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

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

Q = [m (^{1}_{1}H) + (^{3}_{1}H) -2m (^{2}_{1}H)] c^{2}

= [1.007825 + 3.016049 - (2 x 2.014102)] c^{2}

Q = (-0.004433) c^{2}

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

Therefore Q = - (0.004433 x 931.5)

=-4.0334 MeV

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

- The given nuclear reaction is:

^{12}_{6}C + ^{11}_{6}C → ^{20}_{10}Ne + ^{4}_{2}He

It is given that:

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

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

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

The Q-value of this reaction is given as:

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

= [2 x 12.0 – 19.992439 – 4.002603] c^{2}

= (0.004958 c^{2}) u =0.004958 x 931.5 =4.618377 MeV

The positive Q-value of the reaction shows that the reaction is exothermic.

Question 16.

Suppose, we think of fission of a (^{56}_{26}Fe) nucleus into two equal fragments, (^{28}_{13}Al); is the fission energetically possible?

Argue by working out Q of the process. Given m (^{56}_{26}Fe) = 55.93494 u and m (^{28}_{13}Al) = 27.98191 u.

Answer:

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

(^{56}_{26}Fe) à 2 (^{28}_{13}Al)

It is given that:

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

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

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

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

= [55.93494 – (2 x 27.98191)] c^{2}

= (-0.02888 c^{2}) u

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

Therefore, Q = (-0.02888 x 931.5)

=-26.902 MeV

The Q-value of the fission is negative. Therefore, the fission is not possible energetically.

For an energetically-possible fission reaction, the Q-value must be positive.

Question 17.

The fission properties of ^{239}_{94} Pu are very similar to those of ^{235}_{92} U.

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 ^{239}_{94} Pu undergo fission?

Answer:

Given:

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

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

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

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

1 mole of (^{239}_{94}Pu) contains N_{A} atoms.

Therefore mg of (^{239}_{94}Pu) contains ((N_{A})/ (Mass number) x m) atoms

= ((6.023 × 10^{23})/ (239) x1000) =2.52 x10^{24} atoms.

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

E= (E_{av} x 2.52x10^{24})

= (180x2.52x10^{24}) = 4.536x10^{26}MeV

Hence, 4.536x10^{26}MeV is released if all the atoms in 1 kg of pure (^{239}_{94}Pu) undergo fission

**Question 18.**

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ^{235}_{92}U did it contain initially?

Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ^{235}_{92}U and

that this nuclide is consumed only by the fission process.

Answer:

Given:

Half-life of the fuel of the fission reactor, t_{ (1/2) }= 5years

= (5 × 365 × 24 × 60 × 60) s

We know that in the fission of 1 g of (^{235}_{92}U) nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235 g of (^{235}_{92}U) contains (6.023 × 10^{23}) atoms.

1 g (^{235}_{92}U) contains = (6.023x10^{23})/ (235) atoms.

The total energy generated per gram of (^{235}_{92}U) is calculated as:

E = ((6.023x10^{23})/ (235)) x200MeV/g

= (200x6.023x10^{23}x1.6x10^{-19}x10^{6})/ (235) =8.20x10^{10} J/g

The reactor operates only 80% of the time.

Hence, the amount of (^{235}_{92}U) consumed in 5 years by the 1000 MW fission reactor is calculated as:

= (5x80x60x60x365x24x1000x10^{6}) g / (100x8.20x10^{10})

=1538kg

Therefore, Initial amount of (^{235}_{92}U) = (2 × 1538) = 3076 kg

**Question 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

^{2}_{1}H+^{2}_{1}H à ^{3}_{2}He+n+3.27 MeV

Answer:

Given:

The given fusion reaction is:

^{2}_{1}H + ^{2}_{1}H à ^{3}_{2}He+n+3.27MeV

Amount of deuterium, m = 2 kg

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

Therefore, 2.0 kg of deuterium contains= ((6.023x10^{23})/ (2)) x (2000) =6.023x10^{26 }atoms

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

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

E= ((3.27/2) x 6.023x10^{26}) MeV

= ((3.27/2) x 6.023x10^{26}x 1.6x10^{-19 }x 10^{6})

=1.576x10^{14} J

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:

= (1.576x10^{14})/ (100x60x60x24x365)

= (4.9 x10^{4}) years.

**Question 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.)

Answer:

For head on collision, distance between centre of two deuterons = 2 x radius.

= (2 x 2) fm =4 x 10^{-15}m

Charge of each deuteron e= 1.6 x 10^{-19} C

Potential energy = (e^{2})/ (4 π∈_{0}r)

= ((9 x 10^{9}) x (1.6 x 10^{-19})^{2})/ (4 x 10^{-15}) Joule

= (9 x 1.6 x 1.6 x 10^{-14})/ (4 x 1.6 x 1.6^{-16}) keV =390 keV

P.E = 2 x K.E. of each deuteron = 360 keV

K.E. of each deuteron = (360/2) = 180keV.

This is the measure of the height of coulomb or potential barrier.

**Question 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).

Answer:

We have the expression for nuclear radius as:

R = R_{0}A^{1/3}

Where,

R_{0} = Constant.

A = Mass number of the nucleus

Nuclear matter density, ρ = (Mass of the nucleus)/ (volume of nucleus)

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

Therefore ρ= (mA)/ ((4/3) πR^{3}) = (3mA)/ (4π) (R_{0 }A ^{(1/3)}) ^{3}= (3m)/ (4πR_{0}^{3})

Hence, the nuclear matter density is independent of A. It is nearly constant.

**Question 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^{+} + ^{A}_{Z}X à ^{A}_{Z-1}Y + ν

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

Answer:

Given:

Let the amount of energy released during the electron capture process be Q_{1}. The nuclear reaction can be written as:

e^{+} + ^{A}_{Z}X à ^{A}_{(Z-1)}Y + ν + Q_{1} (equation(1))

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

^{A}_{Z}X à ^{A}_{ (Z-1)} Y + e^{+} + ν + Q_{2} (equation (2))

m_{N} (^{ A}_{Z}X) =Nuclear mass of ( ^{A}_{Z}X)

m_{N} (^{A}_{ (Z-1)} Y) = Nuclear mass of (^{A}_{ (Z-1)} Y)

m(^{A}_{Z}X) = Atomic mass of ( ^{A}_{Z}X)

m (^{A}_{ (Z-1)} Y) = Atomic mass of (^{A}_{ (Z-1)} Y)

m_{e} = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q_{1}= [m_{N }(^{A}_{Z}X) + m_{e} -m_{N }(^{A}_{ (Z-1)} Y)] c^{2 }

= [m (^{A}_{Z}X) -Zm_{e }+ m_{e} –m (^{A}_{ (Z-1)} Y) + (Z-1) m_{e}] c^{2}

= [m (^{A}_{Z}X) -m^{ A}_{ (Z-1)} Y] c^{2} (equation (3))

Q-value of the positron capture reaction is given as:

Q_{2}= m_{N }(^{A}_{Z}X) -m_{N }(^{A}_{ (Z-1)} Y) - m_{e}] c^{2}

= [m (^{A}_{Z}X) -Zm_{e }– m (^{A}_{ (Z-1)} Y) + (Z-1) m_{e} - m_{e}] c^{2}

= [m (^{A}_{Z}X) -m^{ A}_{ (Z-1)} Y – 2m_{e}] c^{2} (equation (4))

It can be inferred that if Q_{2} > 0, then Q_{1} > 0; Also, if Q_{1}> 0, it does not necessarily mean that Q_{2} > 0.

In other words, this means that if β^{+ }emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa.

This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

.