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

Class 12th Physics NCERT Solutions for Chapter 13 (Nuclei) are available here. This chapter contains a large number of questions, so, we have provided solutions in two parts. In this part, you will find solutions from question number 13.1 to 13.11. Solutions of other questions are available in part II.  NCERT Solutions for 12th Physics, Chapter 13 - Nuclei are available here. You can also download these solutions in PDF format. Here, you will find solutions from question number 13.1 to 13.11. Some important questions of this chapter are atomic masses and composition of nucleus, discovery of neutron, size of the nucleus, mass – energy relationship, nuclear binding energy, nuclear force radioactivity, Law of radioactive decay, alpha decay, gamma decay, nuclear energy, nuclear Fission-  fusion, nuclear reactor, energy generation in stars, controlled thermonuclear fusion. Questions given in the exercises of this chapter are mostly based on these topics. These topics are important for CBSE 12th Physics board exam 2018.

NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei from question number 13.1 to 13.11 are given below

Question 13.1 (a): Two stable isotopes of lithium 3Li6 and 3Li7 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, 5B10 and 5B11. 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.

Solution 13.1 (a): Given:

Abundance of 3Li6 = 7.5%

Abundance of 3Li7 = 92.5%

Mass of 3Li6 = 6.01512 u

Mass of 3Li7 = 7.016 u

The atomic mass of lithium will depend on the atomic masses of these two isotopes with their abundances as given. This can be calculated by finding the weighted average of the two, as done below

Atomic mass of Li = (6.01512 × 7.5 + 7.016 × 92.5) / 100 = 6.940934 u = 6.941 u

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(b) Given:

Mass of 5B10 = 10.01294 u

Mass of 5B11= 11.00931 u

Atomic mass of B = 10.811 u

The abundances of 5B10 and 5B11 can be calculated by using the technique employed in solution of part (a) of this question.

Let the abundance of 5B10 be y%. Then, the abundance of 5B11 = (100 – y)%

Now calculating weighted average

Atomic mass of boron = [10.01294y + 11.00931(100 – y)] / 100

Or

1081.1 = 10.01294y + 1100.931 – 11.00931y

Simplifying this linear equation and solving further, 0.99636y = 19.831 which gives

y = 19.9%

100 – y = 80.1%.

Thus, abundance of 5B10 = 19.9%

And, abundance of 5B11 = 80.1%.

Question 13.2: The three stable isotopes of neon: 10Ne20, 10Ne21 and 10Ne22 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.

Solution 13.2:

Respective masses and abundances of three isotopes of neon are given

Average atomic mass of Ne = [(19.99 × 90.51) + (20.99 × 0.27) + (21.99 × 9.22)]/100 = 20.1771 u.

Question 13.3: Obtain the binding energy (in MeV) of a nitrogen nucleus 7N14, given m (7N14) = 14.00307 u.

Solution 13.3:

Binding energy of 7N14 = ΔMc2

Mass defect (Δm) = [mass of 7 protons + mass of 7 neutrons – actual mass of N] = [14.11543 – 14.00307] = 0.11236 u

Binding energy = 0.11236 u x 931.5 MeV / u = 104.66 MeV.

Question 13.4: Obtain the binding energy of the nuclei 26Fe56 and 83Bi209 in units of MeV from the following data:

m (26Fe56) = 55.934939 u, m (83Bi209) = 208.980388 u.

Solution 13.4:

Mass defect of Fe = Theoretical value – Practical value

Solving as per previous question,

Mass defect of Fe = 0.528461 u

BE of Fe = 0.528461 × 931.5 = 492.26 MeV

BE per nucleon of Fe = 492.26 / 56 = 8.97 MeV

Similarly,

Mass defect of Bi = 1.760877 u

BE of Bi = 1640.257 MeV

BE per nucleon of Bi = 1640.257 / 209 = 7.84 MeV.

CBSE Class 12th Physics Notes: All Chapter

Question 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 63Cu29 atoms (of mass 62.92960 u).

Solution 13.5:

Given:

Mass of coin = 0.003 kg

Mass of 29Cu63 = 62.9296 u = 1.045261 × 10‒25 kg

No. of Cu atoms in given sample, D = 0.003 / (1.045261 × 10-25) = 2.8701 x 1022 atoms

Mass defect of 1 Cu atom = 0.591935

BE of 1 Cu atom = 0.591935 x 931.5 = 551.387 MeV

So,

BE of D atoms = 2.8701 x 1022 x 551.387 = 1.583 x 1025 MeV = 2.53 x 1012 J.

Question 13.6: Write nuclear reaction equations for

(i) α-decay of 88Ra226

(ii) α-decay of 94Pu242

(iii) β-decay of 15P32

(iv) β-decay of 83Bi210

(v) β+-decay of 6C11

(vi) β+-decay of 43Tc97

(vii) Electron capture of 54Xe120

Solution 13.6:

(i) 88Ra22686Rn222 + 2He4

(ii) 94Pu24292U238 + 2He4

(iii) 15P32 16S32 + e +

(iv) 83Bi21084X210 + e +

(v) 6C115B11 + e+ + v

(vi) 43Tc9742X97 + e+ + v

(vii) 54X120 + ‒1e053X120

Question 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?

Solution 13.7:

Given:

T1/2 = T years

λ = 0.693 / T

(a) N = 0.03125

No = 1

Using the expression

ln N – ln No = ‒ t

Substituting values and solving

-3.466 = [-0.693/T].t

Or t = 5T years.

(b) N = 0.01

No = 1

Similarly

ln N – ln No = ‒ t

Substituting values and solving

t = 6.65T years.

The normal activity of living carbon-containing matter is found to be about

Question 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 6C14 present with the stable carbon isotope 6C12. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity ceases and its activity begins to drop. From the known half life (5730 years) of 6C14, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 6C14 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.

Solution13.8:

Given:

Initial Normal activity R = 15 decays / minute = 0.25 decay / s

T1/2 = 5730 years = 1.807 x 1011 s

Therefore, disintegration constant, λ = 0.693/T1/2 = 0.693 / (1.807 × 1011) = 3.835 × 10-12

Initial No

R = λ No

Or No = 0.25/ λ = 6.52 x 1010 atoms

Specimen activity R’ = 9 decays / minute = 0.15 decay / s

Therefore N of specimen

R’ = N

Or N = 0.15 /(3.835 × 10-12) = 3.91 x 1010 atoms

By law of radioactive decay

ln N – ln No = - λ t

Or t = (24.39 – 24.9)/(‒3.835 x 10-12) = 1.332 × 1011 s =  4224 years

Thus the approximate age of Indus-Valley Civilization from the given sample is 4224 years.

Question 13.9: Obtain the amount of 27Co60 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of27Co60 is 5.3 years.

Solution13.9:

Given,

dN/dt = 8.0 mCi = 8.0 × 3.7 × 107 disintegration/sec

Half life, T =5.3 years = 5.3 × 365 × 24 × 60 × 60 sec = 1.67 × 108 sec

λ = 0.693/T = 0.693/(1.67 × 108) = 4.41 × 10‒9 s‒1

Also, dN/dt = λN ⇒ N = (dN/dt)/λ = (8 × 3.7 × 107)/(4.14×10‒9) = 7.15 × 1016 atoms

Now,

Mass of 6.023 × 1023 atoms of 27Co60 = 60 g

Mass of 7.15 × 1016 atoms of 27Co60 = 7.12 × 10‒6 g.

Question 13.10: The half-life of 38Sr90 is 28 years. What is the disintegration rate of 15 mg of this isotope?

Solution13.10:

Given,

Number of atoms in 90 g of 38Sr90 = 6.023 × 1023

So, number of atoms in 15 g of 38Sr90 = 1020

Rate of disintegration, dN/dt = λ N = (0.693/T1/2) N = [0.693/(28×3.54×107 seconds)] × 0.693 = 7.877 × 1010 Bq.

Question 13.11: Obtain approximately the ratio of the nuclear radii of the gold isotope 79Au197 and the silver isotope 47Ag107.

Solution13.11:

Required ratio = Radius of Au isotope / Radius of Ag isotope = [Ro (197)1/3]/[Ro (107)1/3] = 5.819/4.747 = 1.23.

Download NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 13: Nuclei in PDF format

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