Check important MCQs (with solutions) for CBSE Class 12th Physics Board Exam 2020 (Chapter 13 - Nuclei). Here you will also get important links to access some important articles for the preparation of CBSE 12th board exams 2020. Students preparing for CBSE Class 12th Physics Board Exam 2020 usually ask about important MCQs (Multiple Choice Questions) & here we have provided important questions (with solution), based on Chapter 13 (Nuclei) of Class 12th Physics NCERT textbook.

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**Important MCQs for ****CBSE ****Class 12 Physics**** Board Exam 2020 (Chapter 13 - Nuclei): **

**Chapter 13**

**Nuclei**

**Q1.** The ratio of nuclear radii and nuclear densities of _{26}Fe^{56} and _{92}U^{238} is

(a) 0.671 and 1 respectively

(b) 1 and 0.671 respectively

(c) both equal to 1

(d) both equal to 0.671

**Sol:** (a)

Give, A_{1}= 56, A_{2} = 238

**Q2.** There are 10^{6} radioactive nuclei in a given radioactive element. Its half life is 20 seconds. The number of nuclei will remain left after 10 seconds are:

(a) 3 × 10^{6}

(b) 3 × 10^{5}

(c) 7 × 10^{5}

(d) 2 × 10^{5}

**Sol:** (c)

Given, number of initial nuclei = *N*_{0} = 10^{6}, half life = T = 20 s, time for which decay will occur = *t* = 10 s.

Number of half lives in *t* sec, *n* = t/T = 10/20 = 1/2

Number of nuclei after *n* half lives is given by:

**Q3.** The half life period of a radioactive material if its activity drops to (1/16)^{th} of its initial value in 30 years is

(a) 1.5 years

(b) 3.5 years

(c) 7.5 years

(d) 15 years

**Sol:** (c)

If N is the final number of Nuclei left and N_{0} be the initial number of Nuclei then, according to the question, N = N_{0}/16, when t = 30 years

Number of Nuclei left after *n* half-lives is given by

**Q4.** The half life off _{92}U^{238} against - decay is 4.5 10^{9 }year. The activity of 1 g sample of _{92}U^{238 }is:

(a) 2.2 × 10^{6} Bq

(b) 3.2 × 10^{5} Bq

(c) 1.2 × 10^{4} Bq

(d) 4.1 × 10^{8} Bq

**Sol:** (c)

Given, time of half life = *T* = 4.5 10^{9} years = 4.510^{9}365246060s = 1.42 10^{17}s

We know that,

Number of Nuclei in 1 mole of uranium or 238 g of Uranium‒238 = Avogadro’s number = 6.023 10^{23}

**Q5.** The disintegration energy Q for fission of _{42}M_{o}^{98} into two equal fragments _{21}Sc^{49} by bombarding with a neutron is:

[Given, m (_{42}Mo^{98}) = 97.90541 u, (_{21}Sc^{49}) = 48.95002 u, m_{n} = 1.00867 u]

(a) 341.2 MeV

(b) 944.1 MeV

(c) 5401.2 MeV

(d) None of these

**Sol:** (b)

The disintegration energy in fission of _{42}Mo^{98} is given by

Q = (∆m) X 931MeV = [m (_{42}Mo^{98}) + m_{n} – 2m (_{21}Sc^{49})] X 931 MeV

= [97.90541 + 1.00867 – 2 X 48.95002] X 931 MeV

Q = 1.01404 X 931 = 944.1 MeV

**Q6.** A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme :

If the mass number and atomic number of A are 180 and 72 respectively then the mass number and atomic number for A_{4} are:

(a) 69 and 172 respectively

(b) 115 and 71 respectively

(c) 172 and 69 respectively

(d) None of these

**Sol:** (c)

Emission of 2 alpha particles decreases the mass number by 8 and charge number by 4.

Emission of one β particle increases the charge number by 1 without affecting mass number.

γ emission causes no change.

Therefore, for A_{4}: mass number = 180 ‒ 8 = 172 and atomic number 72 – 4 + 1 = 69

**Q7. **The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an

(a) isobar of parent

(b) isomer of parent

(c) isotone of parent

(d) isotope of parent

**Sol:** (d)

With emission of one α particle decreases the charge number by 2 and emission of two β particles increases he charge number by 2. Therefore, charge number or atomic number after emission of one α particle and two β particles remains the same. Hence the resulting daughter must be an isotope of the parent involving decrease in mass number by 4.

**Q8.** In a sample off radioactive substance, what percentage decays in one mean life time?

(a) 70 %

(b) 63.2 %

(c) 45.8 %

(d) 23%

**Sol:** (b)

Given, t = τ = 1/λ,

⇒ N/N_{0} = e^{-λt} = e^{-1} = 1/e = 1/2.72 = 0.36

Fraction decayed %

**Q9.** The binding energy per nucleon for the parent nucleus is E_{1} and that for the daughter nuclei is E_{2. }Then

(a) E_{1 }> E_{2}

(b) E_{2 }> E_{2 }

(c) E_{1 }= 2E_{2}

(d) E_{2 }= 2E_{1}

**Sol:** (b)

When a heavy nucleus of higher mass number (less stable) splits into two lighter nuclei the daughter nucleus is of less mass number and becomes more stable, having more binding energy per nucleon. Therefore, E_{2 }> E_{1}

**Q10.** Half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially the samples of A and B have equal number of nuclei. Find the ration of remaining number of A and B nuclei after 80 minutes.

(a) 1 : 1

(b) 4 : 1

(c) 1 : 4

(d) 1 : 16

**Sol:** (c)