Ncert Solution Class 12th Physics Chapter – 13 Nuclei Question & Answer

Ncert Solution Class 12th Physics Chapter – 13 Nuclei

Ncert Solution Class 12th Physics Chapter – 13 Nuclei

?Chapter – 13?

✍Nuclei✍

?Question & Answer?

Q .1 (a) Two stable isotopes of lithium 6Li3 and7Li3 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, ¹⁰B₅ and ¹¹B₅ . 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  ¹⁰B₅ and ¹¹B₅.

Answer:

(a) Mass of ⁶Li₃, lithium isotope, m 1 = 6.01512 u

Mass of ⁷Li₃, lithium isotope, m = 7.01600 u 

Abundance of ⁶Li₃ , η₁  = 7.5%

Abundance of ⁷Li₃ , η₂ = 92.5%

The atomic mass of lithium atom is given as:

( b ) Mass of boron isotope ¹⁰B₅ , m = 10.01294 u

Mass of boron isotope ¹¹B₅ , m = 11.00931 u

Abundance of ¹⁰B₅, η₁ = x %

Abundance of ¹¹B₅, η₂ = (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 ¹⁰B₅ 19.89% and that of ¹¹B₅ is 80.11%.

 Q .2 The three stable isotopes of neon: ²⁰Ne₁₀, ²¹Ne₁₀ and ²²Ne₁₀ 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:

Atomic mass of ²⁰Ne₁₀, m₁ = 19.99 u

Abundance of ²⁰Ne₁₀, η₁ = 90.51%

Atomic mass of ²¹Ne₁₀, m₂ = 20.99 u

Abundance of ²¹Ne₁₀, η₂ = 0.27%

Atomic mass of ²²Ne₁₀, m₃ = 21.99 u

Abundance of ²²Ne₁₀, η₃ = 9.22%

The average atomic mass of neon is given as:

Q .3 Obtain the binding energy (in MeV) of a nitrogen nucleus (14/7 N) , given  m(14/7 N) =14.00307 u

Answer:

Atomic mass of nitrogen

(₇N¹⁴)

, m = 14.00307 u

A nucleus of nitrogen 

₇N¹⁴

 contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7m_H + 7m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n= 1.008665 u

∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c²

∴Δm = 0.11236 × 931.5 MeV/c²

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

E_b = Δmc²

Where,

c = Speed of light

∴E_b = 0.11236 × 931.5 

(MeV/c²)×c²

= 104.66334 MeV

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

Q .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:

Atomic mass of

56/26 Fe

, m₁ = 55.934939 u

56/26 Fe

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

Hence, the mass defect of the nucleus, Δm = 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

∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c²

∴Δm = 0.528461 × 931.5 MeV/c²

The binding energy of this nucleus is given as:

E_b1 = Δmc²

Where,

c = Speed of light

∴E_b1 = 0.528461 × 931.5 

MeV/c²×c²

= 492.26 MeV

Average binding energy per nucleon 

Atomic mass of

209/83 Bi

, m₂ = 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:

Δm‘ = 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

∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c²

∴Δm‘ = 1.760877 × 931.5 MeV/c²

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

E_b2 = Δm‘c²

= 1.760877 × 931.5

MeV/c²×c²

= 1640.26 MeV

Average bindingenergy per nucleon = 

Q .5 A given coin has a mass of 3.0 g. Calculate the nuclear energy thatwould be required to separate all the neutrons and protons fromeach other. For simplicity assume that the coin is entirely made of6329Cu atoms (of mass 62.92960 u).

Answer:

Given: The mass of a copper coin 3 g, the atomic mass of copper 29 63 Cu is 62.92960 u

The total numbers of atoms in the coin are given as,

N= N A ×m A

Where, Avogadro number is N A , the number of atoms are N, the mass number of copper A and the mass of a copper coin is m.

By substituting the given values in the above equation, we get

N= 6.022× 10 23 ×3 63 =2.868× 10 22  atoms

The number of protons in the nucleus of 29 63 Cuis 29 and the number of neutrons in the nucleus of 29 63 Cuis 34.

The mass defect of the nucleus of copper atom is given as,

Δ m ′ =29 m p +34 m n −m

Where, the mass of proton is m p , the mass of neutron is m n and the mass defect of the nucleus is Δ m ′ .

By substituting the given values in the above equation, we get

Δ m ′ =29×1.007825+34×1.008665−62.9296 =29.226925+34.29461−62.9296 =0.591935 u

The mass defect of all the atoms present in the coin is given as,

Δm=Δ m ′ ( N )

By substituting the given values in the above equation, we get

Δm=0.591935×2.868× 10 22 =1.69766958× 10 22  u

The binding energy of the nucleus of iron is given as,

E=Δm c 2

Where, the speed of light is c and the binding energy is E.

By substituting the given values in the above equation, we get

E=1.69766958× 10 22 c 2 ×931.5  MeV/ c 2 =1.584× 10 25 ×1.6× 10 −13  J =2.535× 10 12  J

Thus, the amount of energy required to separate all the neutrons and protons from the given coin is 2.535× 10 12  J.

Q .6 Write nuclear reaction equations for
(i) α-decay of 22688 Ra
(ii) α-decay of 24294 Pu
(iii) β–-decay of 3215 P
(iv) β–-decay of 21083 Bi
(v) β+-decay of 116 C
(vi) β+-decay of 9743 Tc
(vii) Electron capture of 12054 Xe

Answer:

• In alpha decay the mass number is reduced by four and the atomic number by two R 86 224 a.

• P 92 238 u

• In beta minus decay the P 15 32 → S 16 32 + β −1 0

• B 83 210 i→ P 84 210 o+ β −1 0

• After beta plus decay C 6 11 → B 5 11 + β 1 0 +

• After beta plus decay T 43 97 c→ M 42 97 o

• When the electron is captured by the xenon then

•  X 54 12 e+ e −1 → I 53 120

Q .7 A radioactive isotope has a half-life of T years. How long will it takethe activity to reduce to a) 3.125%, b) 1% of its original value?

Answer:

a) Given: The half life of the radioactive isotope is T years and the reduction in the activity is 3.125%.

According to the law of radioactive decay,

N= N 0 e −λt N N 0 = e −λt

Where, N is the remaining number of molecules, N 0 is the initial number of atoms, λ is the decay constant and t is the time.

By substituting the given values in the above expression, we get

3.125 100 = e −λt 1 32 = e −λt λt= log e ( 32 ) t= 3.4657 λ

We know that,

λ= 0.693 T

By substituting the given value in the above expression, we get

t= 3.466 0.693 T =5T years

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

b) Given: The reduction in the activity is 1%.

According to the law of radioactive decay,

N= N 0 e −λt N N 0 = e −λt

By substituting the given values in the above expression, we get

1 100 = e −λt 1 100 = e −λt λt= log e ( 100 ) t= 4.6052 λ

We know that,

λ= 0.693 T

By substituting the given value in the above expression, we get

t= 4.0652 0.693 T =6.65T years

Thus, the isotope will take about 6.65T years to reduce to 1% of its original value.

 Q 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 14/6C  present with the stable carbon isotope 12/6C  . 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 14/6C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 14/6C 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.

Answer:

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

14/6C

T 1/2

= 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, λand time, t as:

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

Q .9 Obtain the amount of  60/27 C₀ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of  60/27 C₀ is 5.3 years.

Answer:

The strength of the radioactive source is given as:

Where,

N = Required number of atoms

Half-life of

60/27 C₀

T 1/2

 = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10⁸ s

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

dN/dt = λ N

Where, λ 

For

₂₇C0⁶⁰

Mass of 6.023 × 10²³ (Avogadro’s number) atoms = 60 g

∴Mass of 

7.133×10¹⁶

atoms 

Hence, the amount of 

₂₇C0⁶⁰

 necessary for the purpose is 7.106 × 10⁻⁶ g.

Q .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 × 10⁸ s

Mass of the isotope, m = 15 mg

90 g of 

90/38 Sr

atom contains 6.023 × 10²³ (Avogadro’s number) atoms.

Therefore, 15 mg of 

90/38 Sr

 contains:

Rate of disintegration, 

dN/dt = λ N

Where,

λ = Decay constant 

Hence, the disintegration rate of 15 mg of the given isotope is
7.878 × 10¹⁰ atoms/s.

 Q 11. Obtain approximately the ratio of the nuclear radii of the gold isotope 19779 Au and the silver isotope 10747 Ag ..

Answer:

The mass number of gold isotope A 79 197 uis 197 and the mass number of silver isotope A 47 107 g is 107.

The radius of an isotope is directly proportional to its mass number.

R∝ A 1 3

Where, R is the radius of the nucleus and A is the mass number.

The ratio of radii of two nuclei is given as,

R Au R Ag = ( A Au A Ag ) 1 3

By substituting the given values in above equation, we get

R Au R Ag = ( 197 107 ) 1 3 =1.2256 ≈1.23

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

 Q .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(226/88 Ra)  = 226.02540 u, m(222/86 Rn) = 222.01750 u,m(220/86 Rn)= 220.01137 u, m(216/84 Po) = 216.00189 u.

Answer: Alpha particle decay of 226/88Ra 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) c²

Where,

c = Speed of light

It is given that:

Q-value = [226.02540 − (222.01750 + 4.002603)] u c² 
= 0.005297 u c²

But 1 u = 931.5 MeV/c²

∴ Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle  =

= 222/226×4.94 = 4.85 MeV

Q .13 The radionuclide ¹¹C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

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

Answer:

The given nuclear reaction is:

Atomic mass of 

m(11/6 )C

= 11.011434 u

Atomic mass of 

m(11/6 B) = 11.009305

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the 

(11/6 )C

 nucleus is given as:

Where,

m_e = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 m_e in the case of

¹¹C

and 5 m_e in the case of¹¹B

Hence, equation (1) reduces to:

∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c²

= (0.001033 c²) u

But 1 u = 931.5 Mev/c²

∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

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

Q .14 The nucleus 2310 Ne decays by β–emission. Write down the β-decay equation and determine the maximum kinetic energy of theelectrons emitted. Given that:m (2310 Ne ) = 22.994466 um (2311 Na ) = 22.089770 u.

Answer:

 Q .15 The Q value of a nuclear reaction A+b⟶C+d is defined by Q = [ m_A+ m_b− m_C− m_d]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.

Answer:

The given nuclear reaction is:

It is given that:

Atomic mass m(1/1H) = 1.007825 u

Atomic mass m(1/3H) = 3.016049 u

Atomic mass m(1/2H)=2.014102u

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

Q = (- 0.00433 c²)u

But 1 u = 931.5 MeV/c²

=[1.007825 +3.016049-2×2.014102]c²

Q =-0.00433×931.5=-4.0334 MeV

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

 Q .16 Suppose, we think of fission of a 5626Fe nucleus into two equalfragments, 2813 Al . Is the fission energetically possible? Argue byworking out Q of the process. Given m ( 5626Fe ) = 55.93494 u andm ( 2813 Al ) = 27.98191 u.

Answer:

Q .17 The fission properties of 23994 Pu are very similar to those of 23592 U . Theaverage energy released per fission is 180 MeV. How much energy,in MeV, is released if all the atoms in 1 kg of pure 23994 Pu undergofission?

Answer:

Given: The average energy released per fission of P 94 239 u is 180 MeV and the amount of P 94 239 u is 1 kg.

239 gof P 94 239 u contains 6.023× 10 23  atoms.

Let N be the number of atoms present in 1 kgof P 94 239 u.

The number of atoms contain by 1 kg of P 94 239 u is given by,

N= 6.023× 10 23 239 ×1000 =2.52× 10 24  atoms

Let E t be the total energy released in the fission.

E t =E×N

Where, the average energy released per fission of P 94 239 uis E.

By substituting the given values in above equation, we get.

E t =180×2.52× 10 24 =4.536× 10 26  MeV

Thus, the energy released by the fission of 1 kgof P 94 239 uis 4.536× 10 26  MeV.

Q .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:

Half life of the fuel of the fission reactor, 

^t1/2 = 5

years

= 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 200 MeV.

1 mole, i.e., 235 g of 

235/92 U

contains 6.023 × 10²³ atoms.

∴1 g 

235/92 U

 contains

6.023×10²³/235 atoms.

The total energy generated per gram of

235/92 U

is calculated as:

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:

∴Initial amount of 

235/92 U

= 2 × 1538 = 3076 kg

Q .19 How long can an electric lamp of 100W be kept glowing by fusion of 2Kg of deuterium? Take fusion reaction is 1H2 + 1H2 = 2He4 + n+ 3.27MeV

Answer:

The given fusion reaction is:

Amount of deuterium, m = 2 kg

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

And 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:

Q .20 Calculate the height of the potential barrier for a head on collisionof two deuterons. (Hint: The height of the potential barrier is givenby the Coulomb repulsion between the two deuterons when theyjust touch each other. Assume that they can be taken as hardspheres of radius 2.0 fm.)

Answer:

Given, the radius of deuteron nucleus is 2 fm.

Let the distance between the centres of two deuterons be d.

Formula for the distance dbetween the two deuterons is, d=radius of 1 deuteron +radius of 2 deuteron

Substitute the values in the above equation.

d=2× 10 −15 +2× 10 −15 =4× 10 −15  m

Charge on a deuteron nucleus is e=1.6× 10 −19 C

Potential energy of the two-deuteron system is given by,

V= e 2 4π ε ° d

Here, ε 0 is the permittivity of free space..

Substituting the values in the above equation, we get:

V= e 2 4π ε ° d = ( 1.6× 10 −19 ) 2 4×π×8.85× 10 −12 ×4× 10 −15 =360 keV

Thus, the height of the potential barrier of the two-deuteron system is 360 keV.

Q .21 From the relation R = R₀A¹/³, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Answer:

We have the expression for nuclear radius as:

R = R₀A¹/³

Where,

R₀ = Constant.

A = Mass number of the nucleus

Nuclear matter density, 

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

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

Q .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).

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

Answer: Let the amount of energy released during the electron capture process be Q₁. The nuclear reaction can be written as:

Q .23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are ²⁴₁₂Mg (23.98504u), ²⁵₁₂Mg(24.98584u) and ²⁶₁₂Mg (25.98259u). The natural abundance of ²⁴₁₂Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.

Answer: Let the abundance of ²⁵₁₂Mg be x%

The abundance of ²⁶₁₂Mg = (100 – 78.99 -x)%

= (21.01 – x)%

The average atomic mass of magnetic

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

24.312  x 100= 1894.57 + 24.98584x + 545. 894 – 25.98259x

2431.2 = 2440.46 – 0.99675x

0.99675x = 2440.46 – 2431.2

0.99675x =  9.26

x = 9.26/0.99675

x = 9.290%

Abundance of ²⁵₁₂Mg is 9.290%

The abundance of ²⁶₁₂Mg = (21.01 – x)%

= (21.01 -9.290 )% = 11.71%

Q . 24 The neutron separation energy is defined as the energy required toremove a neutron from the nucleus. Obtain the neutron separationenergies of the nuclei 4120Ca and 2713Al from the following data:m(4020Ca ) = 39.962591 um(4120Ca ) = 40.962278 um(2613Al ) = 25.986895 um(2713Al ) = 26.981541 u

Answer:

When the nucleons are separated from ₂₀Ca⁴¹ → ₂₀ Ca⁴⁰ + ₀n¹

Mass defect, Δm = m (₂₀ Ca⁴⁰) + m_n – m(₂₀ Ca⁴¹)

= 39.962591 + 1.008665 – 40.962278

= 0.008978 amu

Neutron separation energy = 0.008978  x 931 MeV = 8.362 MeV

Similarly, ₁₃Al²⁷ → ₁₃ Al²⁶ + ₀n¹

Mass defect, Δm = m (₁₃ Al²⁶) + m_n – m(₁₃Al²⁷)

= 25.986895 + 1.008665 – 26.981541

= 26.99556 – 26.981541 = 0.014019 amu

Neutron separation energy = 0.014019 x 931 MeV

= 13.051 MeV

Q . 25 A source contains two phosphorous radionuclides ³²₁₅P (T_1/2 = 14.3d) and ³³₁₅P(T_1/2 = 25.3d). Initially, 10% of the decays come from ³³₁₅ P. How long one must wait until 90% do so?

Answer:

Let initially there be N₁ atoms of

32/15P

and N₂ atoms of

32/15P

and let their  decay constants be

λ₁

and

λ₂

respectively

Since initially the activity of

32/15P

is 1/9 times that of

32/15P

we have

(i) Let after time t the activity of

32/15P

be 9 times that of

32/15P

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

where

λ_{2 = 0.048/ day}

and

λ_{1 = 0.027/ day}

t comes out to be 208.5 days

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

223/88 Ra →209/82 Pb + 14/6 C

223/88 Ra →219/86 Rn + 4/2 He

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

Answer:

Take a 

14/6 C

 emission nuclear reaction:

223/88 Ra →209/82 Pb + 14/6 C

We know that:

Mass of 

223/88 Ra

m₁ = 223.01850 u

Mass of 

209/82 Pb

m₂ = 208.98107 u

Mass of

14/6 C

, m₃ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = (m₁ − m₂ − m₃) c²

= (223.01850 − 208.98107 − 14.00324) c²

= (0.03419 c²) u

But 1 u = 931.5 MeV/c²

∴Q = 0.03419 × 931.5

= 31.848 MeV

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

Now take a 

4/2 He

emission nuclear reaction:

223/88→229/86 Rn + 4/2 He

We know that:

Mass of 

223/88 Ra

m₁ = 223.01850

Mass of 

219/82 Rn

m₂ = 219.00948

Mass of

4/2 He

, m₃ = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m₁ − m₂ − m₃) c²

= (223.01850 − 219.00948 − 4.00260) C²

= (0.00642 c²) u

= 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.

Q . 27 Consider the fission of 23892U by fast neutrons. In one fission event,no neutrons are emitted and the final end products, after the betadecay of the primary fragments, are 14058Ce and 9944Ru. Calculate Qfor this fission process. The relevant atomic and particle massesarem(23892U ) =238.05079 um(14058Ce ) =139.90543 um(9944Ru ) = 98.90594 u

Answer: Given: The mass of U 92 238 is 238.05079 u, the mass of C 58 140 e is 139.90543 u and the mass of R 44 99 u is 98.90594 u.

Since, in the fission of U 92 238 , 10 β − particles decay from parent nucleus.

The nuclear reaction is given as,

U 92 238 + n 0 1 → C 58 140 e+ R 44 99 u+10 e −1 0

The Q value of the reaction process are given as,

Q=( m ′ 1 + m 4 – m ′ 2 – m ′ 3 −10 m e ) c 2 (1)

Where, the atomic mass of nuclei of U 92 238 , C 58 140 e, R 44 99 u are m ′ 1 , m ′ 2 , m ′ 3 respectively and the mass of a neutron is m 4 .

The atomic masses of nuclei are given as,

[ m ′ 1 = m 1 −92 m e m ′ 2 = m 2 −58 m e m ′ 3 = m 3 −44 m e ](2)

By substituting the values of atomic masses of nuclei from the equation (2) to equation (1), we get

Q=( m 1 −92 m e + m 4 − m 2 −58 m e − m 3 +44 m e −10 m e ) c 2

Where, the atomic mass of U 92 238 , C 58 140 e, R 44 99 u are m 1 , m 2 , m 3 respectively and the mass of electron is m e .

Since, 1 u=931.5 MeV/ c 2

By substituting the given values in the above expression, we get

Q=( m 1 + m 4 – m 2 – m 3 ) c 2 =( 238.0507+1.008665−139.90543−98.90594 ) c 2 =( 0.247995 c 2 ) u =231.1 MeV

Hence, the Qvalue of fission process is 231.1 MeV.

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

(a) Calculate the energy released in MeV in this reaction from the
data:
m(²₁H )=2.014102 u
m(³₁H ) =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(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Answer:

The Q value is given as

Q = Δm x 931 MeV

= (m (₁H²) + m(₁H³) – m(₂He⁴) – m_n) x 931

=( 2.014102 + 3.016049- 4.002603-1.00867) x 931

Q = 0.0188 x 931 = 17.58 MeV

(b) Repulsive potential energy of two nuclei when they touch each other is

= e²/4πϵ₀(2r)​

Where ε₀ = Permittivity of free space

4Πϵ0 ​= 9 x 10⁹ Nm²C^-2

= (23.04 x 10^-29)/(4 x 10^-15)

= 5. 76 x 10^-14 J

Kinetic energy needed to overcome the coulomb repulsion between the two nuclei is

K.E = 5. 76 x 10^-14 J

K.E= 2 x (3/2) KT

K is the Boltzmann constant = 1.38 x 10^-23

T = K.E/3K = 5. 76 x 10^-14/(3 x 1.38 x 10^-23)

= 5. 76 x 10^-14/(4.14 x 10^-23)

= 1.3913 x 10⁹ K

Q . 29 Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that

m (¹⁹⁸Au) = 197.968233 u

m (¹⁹⁸Hg) =197.966760 u

Answer: It can be observed from the given γ-decay diagram that γ₁ decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ₁-decay is given as:

E₁ = 1.088 − 0 = 1.088 MeV

hν₁= 1.088 × 1.6 × 10⁻¹⁹ × 10⁶ J

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

ν₁ = Frequency of radiation radiated by γ₁-decay

It can be observed from the given γ-decay diagram that γ₂ decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ₂-decay is given as:

E₂ = 0.412 − 0 = 0.412 MeV

hν₂= 0.412 × 1.6 × 10⁻¹⁹ × 10⁶ J

Where,

ν₂ = Frequency of radiation radiated by γ₂-decay

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

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

E₃ = 1.088 − 0.412 = 0.676 MeV

hν₃= 0.676 × 10⁻¹⁹ × 10⁶ J

Where,

ν₃ = Frequency of radiation radiated by γ₃-decay

Mass of 

m(198/78 Au)

= 197.968233 u

Mass of 

m(198/80 Hg)

= 197.966760 u

1 u = 931.5 MeV/c²

Energy of the highest level is given as:

β₁ decays from the 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the β₁ particle = 1.3720995 − 1.088

= 0.2840995 MeV

β₂ decays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the β₂ particle = 1.3720995 − 0.412

= 0.9600995 MeV

Q 30. Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within the sun and b) the fission of 1.0 kg of ²³⁵U in a fission reactor.

Answer: Amount of hydrogen, m = 1 kg = 1000 g

(a) 1 mole, i.e., 1 g of hydrogen contains 6.023 x 10²³ atoms. Therefore, 1000 g of hydrogen contains 6.023 x 10²³  x 1000 atoms.

In Sun, four hydrogen nuclei fuse to form a helium nucleus and will release energy of 26 MeV

Energy released by the fusion of 1 kg of hydrogen,

E₁ = (6.023 x 10²³  x 1000 x 26)/4

E₁= 39.16 x 10²⁶ MeV

(b) Amount of  ²³⁵U = 1 kg = 1000 g

1 mole, i.e., 235 g of uranium contains 6.023 x 10²³ atoms. Therefore, 1000 g of uranium contains (6.023 x 10²³  x 1000)/235 atoms.

The energy released during the fission of one atom of ²³⁵U = 200 MeV

Energy released by the fission of 1 Kg of ²³⁵U,

E₂= (6.023 x 10²³  x 1000 x 200)/235

E₂ = 5.1 x 10²⁶ MeV

(E₁/E₂) = (39.16 x 10²⁶/5.1 x 10²⁶)

= 7.67

The energy released in the fusion of hydrogen is 7.65 times more than the energy released during the fission of Uranium.

Q 13.31. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten per cent of which was to be obtained from nuclear power plants. Suppose we are given that, on average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ²³⁵U to be about 200MeV.

Answer:

Electric power to be generated = 2 x 10⁵ MW = 2 x 10⁵ x 10⁶ J/s = 2 x 10¹¹ J/s

10 % of the amount is obtained from the nuclear power plant

P₁ = (10/100) x 2 x 10¹¹ x 60 x 60 x 24 x 365 J/year

Heat energy released during per fission of a ²³⁵U nucleus, E = 200 MeV

Efficiency of the reactor = 25%

The amount of energy converted to electrical energy in fission is (25/100) x 200 = 50 MeV

= 50 x 1.6 x 10^-19 x 10⁶  = 80 x 10^-13 J

Required number of atoms for fission per year

= [(10/100) x 2 x 10¹¹ x 60 x 60 x 24 x 365]/80 x 10^-13

= 788400 x 10²³ atoms

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

Mass of 6.023 x 10²³ atoms of ²³⁵U = 235 x 10^-3 kg

Mass of 788400 x 10²³  atoms of ²³⁵U

=[ (235 x 10^-3)/(6.023 x 10²³) ] x 78840 x 10²⁴ 

= 3. 076 x 10⁴ Kg

The mass of uranium needed per year is 3.076 x 10⁴ Kg