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 12^{th} 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

_{3}Li

^{6}and

_{3}Li

^{7}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,

_{5}B

^{10}and

_{5}B

^{11}. 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}B

_{5}and

^{11}B

_{5}.

**Solution 13.1 ( a):** Given:

Abundance of _{3}Li^{6} = 7.5%

Abundance of _{3}Li^{7} = 92.5%

Mass of _{3}Li^{6} = 6.01512 u

Mass of _{3}Li^{7}_{ }= 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

**NCERT Exemplar Problems: CBSE Class 12 Physics – All Chapters**

**( b)** Given:

Mass of _{5}B^{10} = 10.01294 u

Mass of _{5}B^{11}= 11.00931 u

Atomic mass of B = 10.811 u

The abundances of _{5}B^{10} and _{5}B^{11} can be calculated by using the technique employed in solution of part (a) of this question.

Let the abundance of _{5}B^{10} be y%. Then, the abundance of _{5}B^{11} = (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 _{5}B^{10} = 19.9%

And, abundance of _{5}B^{11} = 80.1%.

**Question 13.2:** The three stable isotopes of neon: _{10}Ne^{20}, _{10}Ne^{21} and _{10}Ne^{22 }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 _{7}N^{14}, given *m* (_{7}N^{14}) = 14.00307 u.

**Solution 13.3:**

Binding energy of _{7}N^{14} = Δ*Mc*^{2}

Mass defect (Δ*m*) = [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 _{26}Fe^{56 }and_{ 83}Bi^{209} in units of MeV from the following data:

*m* (_{26}Fe^{56}) = 55.934939 u, *m* (_{83}Bi^{209}) = 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 ^{63}Cu_{29} atoms (of mass 62.92960 u).

**Solution 13.5:**

Given:

Mass of coin = 0.003 kg

Mass of _{29}Cu^{63} = 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 10^{22} 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 10^{22} x 551.387 = 1.583 x 10^{25 }MeV = 2.53 x 1012 J.

**Question 13.6:** Write nuclear reaction equations for

(*i*) α-decay of _{88}Ra^{226}

(*ii*) α-decay of _{94}Pu^{242}

(*iii*) β^{–}-decay of _{15}P^{32}

(*iv*) β^{–}-decay of _{83}Bi^{210}

(*v*) β^{+}-decay of _{6}C^{11}

(*vi*) β^{+}-decay of _{43}Tc^{97}

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

**Solution 13.6:**

(*i*) _{88}Ra^{226} → _{86}Rn^{222} + _{2}He^{4}

(*ii*) _{94}Pu^{242} → _{92}U^{238} + _{2}He^{4}

(*iii*) _{15}P^{32 }→_{16}S^{32} + e^{‒} + *ṽ*

(iv) _{83}Bi^{210} → _{84}X^{210} + e^{‒} + *ṽ*

(*v*) _{6}C^{11} → _{5}B^{11} + e^{+} + *v*

(*vi*) _{43}Tc^{97} → _{42}X^{97} + e^{+} + *v*

(*vii*) _{54}X^{120} + _{‒1}*e*^{0} → _{53}X^{120}

**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:

*T*_{1/2 }= T years

λ = 0.693 / T

(*a*) N = 0.03125

*N*_{o} = 1

Using the expression

ln *N* – ln *N*_{o} = ‒ *t*

Substituting values and solving

-3.466 = [-0.693/T].t

Or *t *= 5T years.

(*b*) *N* = 0.01

*N*_{o} = 1

Similarly

ln *N *– ln *N*o = ‒ 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 _{6}C^{14 }present with the stable carbon isotope _{6}C^{12}. 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 _{6}C^{14}, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of _{6}C^{14} 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

T_{1/2} = 5730 years = 1.807 x 10^{11} s

Therefore, disintegration constant, λ = 0.693/T_{1/2} = 0.693 / (1.807 × 10^{11}) = 3.835 × 10^{-12}

Initial N_{o}

R = λ N_{o}

Or *N*_{o} = 0.25/ λ = 6.52 x 10^{10 }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 10^{10} atoms

By law of radioactive decay

ln *N* – ln *N*_{o} = - λ t

Or t = (24.39 – 24.9)/(‒3.835 x 10^{-12}) = 1.332 × 10^{11} 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 _{27}Co^{60} necessary to provide a radioactive source of 8.0 mCi strength. The half-life of_{27}Co^{60} is 5.3 years.

**Solution13.9:**

Given,

d*N*/d*t* = 8.0 mCi = 8.0 × 3.7 × 10^{7} disintegration/sec

Half life, T =5.3 years = 5.3 × 365 × 24 × 60 × 60 sec = 1.67 × 10^{8} sec

λ = 0.693/T = 0.693/(1.67 × 10^{8}) = 4.41 × 10^{‒9} s^{‒1}

Also, d*N*/d*t* = λN ⇒ N = (d*N*/d*t*)/λ = (8 × 3.7 × 10^{7})/(4.14×10^{‒9}) = 7.15 × 10^{16} atoms

Now,

Mass of 6.023 × 10^{23} atoms of _{27}Co^{60} = 60 g

Mass of 7.15 × 10^{16} atoms of _{27}Co^{60} = 7.12 × 10^{‒6} g.

**Question 13.10: **The half-life of _{38}Sr^{90 }is 28 years. What is the disintegration rate of 15 mg of this isotope?

**Solution13.10: **

Given,

Number of atoms in 90 g of _{38}Sr^{90} = 6.023 × 10^{23}

So, number of atoms in 15 g of _{38}Sr^{90} = 10^{20}

Rate of disintegration, dN/dt = λ N = (0.693/T_{1/2}) N = [0.693/(28×3.54×10^{7} seconds)] × 0.693 = 7.877 × 10^{10} Bq.

**Question 13.11**: Obtain approximately the ratio of the nuclear radii of the gold isotope _{79}Au^{197} and the silver isotope _{47}Ag^{107}.

**Solution13.11: **

Required ratio = Radius of Au isotope / Radius of Ag isotope = [R_{o} (197)^{1/3}]/[R_{o} (107)^{1/3}] = 5.819/4.747 = 1.23.

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