# NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei (Part II)

NCERT Solutions for Class 12th Physics - Chapter 13 (Nuclei) are available here from question number 13.11 to 13.22. Solutions from question number 13.1 to 13.10 are available in part I. These questions are important for CBSE board exam and other competitive exams.

Created On: Jun 7, 2017 17:45 IST NCERT Solutions for 12th Physics, Chapter 13 - Nuclei are available here. In this article, you will find solutions from question number 13.12 to 13.22. This article is a continuation of NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei (Part I) where solutions from question number 13.1 to 13.11 are available. These solutions are frequently asked in CBSE board exam 2018 and competitive exams (like JEE Main, WBJEE, NEET etc).

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei from question number 13.12 to 13.22 are given below

Question 13.12: Find the Q-value and the kinetic energy of the emitted α-particle in the α- decay of (a) 88Ra226 and (b) 86Rn220.

Given m (88Ra226) = 226.02540 u, m (86Rn222) = 222.01750 u, m (86Rn222) = 220.01137 u, m (84Po216) = 216.00189 u.

Solution 13.12

(a) Nuclear reaction:

88Ra22686Rn222 + 2He4

Q = (MRa MRnMHe).c2

= (5.297 × 10-3) × (931.5) = 4.934 MeV

Kinetic energy of emitted alpha particle:

The energy liberated in the nuclear reaction displays in the form of kinetic energy of the particles of products.

Kinetic energy of products

(MV2/2) + (mv2/2) = 4.93 MeV …(1)

CBSE Class 12th Physics Notes: All Chapter

Where,

M = mass of radon nuclei

m = mass of helium nuclei

V = velocity of radon nuclei

v = velocity of helium nuclei

To keep the momentum conserve

MV = mv

Or

M / m = v / V…………………………….(2)

This implies nothing but the simple fact that the most of the kinetic energy

(or Q calculated above) will be retained by the helium nuclei, eq. (2) pointing that it will travel with much large velocity as compared to the velocity of the radon nuclei.

Dividing eq. (1) throughout by V2 and making substitutions from eq. (2)

(M2/m) + M = 9.86 / V2

Or

V2 = 7.86 × 10-4

KE of Rn = MV2 / 2 = 0.087 MeV

Therefore, KE of alpha particle = 4.93 – 0.087 = 4.85 MeV.

(b) Nuclear reaction:

86Rn22084Po216 + 2He4

Calculating Q in a similar fashion, Q = 6.41 MeV

Kinetic energy of emitted alpha particle:

As solved in part (a) of this question

KE of products

[(MV)V / 2] + [(mv)v / 2] = 6.41

where

M = mass of polonium nuclei

m = mass of helium nuclei

V = velocity of polonium nuclei

v = velocity of helium nuclei

By the same arguments and solving technique used in part (a) of this question, V comes to be

V = 0.033

KE of Po = MV2 / 2 = 0.117 MeV

Therefore KE of alpha particle = 6.41 – 0.117 = 6.29 MeV

CBSE Class 12th Physics Notes: All Chapter

Question 13.13: The radionuclide 11C decays according to

11C65B11 + e+ + v: half life = 20.3 minute.

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (6C11) = 11.011434 u and m (6B11) = 11.009305 u.

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

Solution 13.13:

Nuclear Reaction:

6C115B11 + e+ + v

Δ m = [m (6C11) ‒ m (5B11) ‒ 2 me] = 0.001033 u

Now, 1 u = 931 MeV

Q = 0.001033 × 931 MeV = 0.961 MeV, which is maximum energy emitted by positron.

This Q has energy distributed as

Q = Ed + Ee + Ev

For constant momentum, the energy carried by daughter nuclei and neutrino is nearly zero. The positron carries maximum energy.

Hence max Ee ≈ Q.

Question 13.14: The nucleus 10Ne23 decays of β emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (10Ne23) = 22.994466 amu, m (11Na23) = 22.989770 amu.

Solution 13.14:

Nuclear Reaction:

10Ne23 11Na23 + e- + n + Q

Q = [mass of Ne – mass of Na – mass of electron] × c2

Using atomic masses

Q = [mass of Ne – mass of Ma].c2 = 4.37 MeV

For the reasons argued in previous questions, the maximum kinetic energy of electron

Maximum Ee = Q = 4.37 MeV.

Question 13.15: The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mbmC md]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

(i) 1H1 + 1H31H2 + 1H2

(ii) 6C12 + 6C12   → 10Ne20 + 2He4

Atomic masses are given to be

m(1H2) = 2.014102 u; m(1H3) = 3.016049 u; m(6C12) = 12.000000 u; m(10Ne20) = 19.992439 u

Solution 13.15:

A + b → C + d

Q = [mA + mbmC md]c2

(i) Nuclear Reaction:

1H1 + 1H31H2 + 1H2

Q = [-4.33 X 10-3] × [931.5] = -4.03 MeV

Reaction is endothermic. This much energy has to be supplied.

(ii) Nuclear Reaction:

6C12 + 6C1210Ne20 + 2He4

Q = [4.958 x 10-3] × [931.5] = 4.62 MeV

Question 13.16: Suppose, we think of fission of a 26Fe56 nucleus into two equal fragments 13Al28. Is the fission energetically possible? Argue by working out Q of the process.

Solution 13.16:

Given, m (26Fe56) = 55.93494 u, m (13Al28) = 27.98191 u.

Nuclear Reaction:

26Fe5613Al28 + 13Al28 + Q

Q = [mass of Fe – mass of aluminium – mass of aluminium].c.c = [-0.0288].[931.5] = ‒26.9 MeV

The reaction being endothermic is not possible.

Question 13.17:  The fission properties of 84Pu239 are very similar to those of 92U235. 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 94Pu239 undergo fission.

Solution 13.17:

Given:

Energy released per fission = 180 MeV

1 g of Pu contains = Avogadro’s No. / 239

1 kg of Pu contains, N = 2.5197 x 1024 atoms

1 atom releases = 180 MeV

Therefore,

N atoms release = 4.536 x 1026 MeV.

Question 13.18:  A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92U235 did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92U235 and that this nuclide is consumed only by the fission process.

Solution 13.18:

Energy generated per gram of U = (6 x 1023 x 200 x 1.6 x 10-13)/235

The amount of U consumed in 5 years with 80% on-time

= (5 x 0.8 x 3.154 x 1016 x 235) / (1.2 x 1.6 x 1013) = 1544 kg

Thus the initial amount = 3088 kg

Question 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

1H2 + 1H22He3 + n + 3.27 MeV

Solution 13.19:

Nuclear Reaction:

1H2 + 1H22He3 + n + 3.27 MeV

No. of atoms in 2 kg deuterium

S = (Avogadro’s No. × 2000)/2 = 6.023 x 1026 atoms

Energy of 1 atom = 3.27 / 2 = 1.635 MeV

Therefore

Energy of S atoms = 9.85 x 1026 MeV = 1.576 x 1014 J.

We know

Power = Energy / Time

Hence,

Time = 1.576 x 1012 s = About 4.9 x 104 years.

Question 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.)

Solution 13.20:

Given:

R = 2 fm

d = R + R = 4 fm

Height of the potential barrier, V = e2/(4πϵod)

Substituting values yields

V = 5.76 x 10-14 J = 360 keV.

Question 13.21: 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).

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 + zXAz‒1YA + v

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

Solution 13.21:

We know

Nuclear mass density = mass / volume …(1)

Now Mass ∝ A

And Volume ∝ R3

Since R ∝ A1/3

Therefore

Volume ∝ A

When putting these proportional values in eq. (1), A gets cancelled. Hence proved!

Question 13.22: 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+ + zXAz‒1YA + v

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

The two processes:

For positron capture,

ZXAz‒1YA + e+ + v + Q1

For electron capture,

e + ZXAz‒1YA + v + Q2

Calculating Q1:

Q1 = [mN(ZXA) – mN (z‒1YA) ‒ 2me] c2

Q2 = [mN(ZXA) + me ‒ mN (z‒1YA)]c2

From the above calculations, it is obvious that, if Q1 > 0 then Q2 > 0

But if Q2 > 0, then it can’t be said that Q1 > 0

Hence proved!

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