Category: Nuclear Physics

Q: A radioactive sample S₁ having an activity of 5 μCi has twice the number of nuclei as another sample S₂ which has an activity of 10 μCi. The half lives of S₁ and S₂ can be

(A) 20 yr and 5 yr, respectively

(B) 20 yr and 10 yr, respectively

(C) 10 yr each

(D) 5 yr each

Ans: (A)

Sol: Activity = λ N

According to question ;

$\large \lambda_1 N_1 = \frac{1}{2}( \lambda_2 N_2 )$

$\large \frac{\lambda_1}{\lambda_2} = \frac{N_2}{2 N_1}$

Since $\lambda \propto \frac{1}{T}$ ; (Where T = half life)

$\large \frac{T_1}{T_2} = \frac{2 N_1}{N_2}$

$\large \frac{T_1}{T_2} = 4 $ ; (Given , N₁ = 2 N₂)

Q: In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is

(A) E (²³⁶U₉₂ ) > E (¹³⁷I₅₃ ) + E (⁹⁷Y₃₉ ) + 2E (n)

(B) E (²³⁶U₉₂) < E (¹³⁷I₅₃ ) + E (⁹⁷Y₃₉) + 2E (n)

(C) E (²³⁶U₉₂ ) < E (¹⁴⁰Ba₅₆ ) + E (⁹⁴Kr₃₆) + 2E (n)

(D) E (²³⁶U₉₂) < E (¹⁴⁰Ba₅₆ ) + E (⁹⁴Kr₃₆) + E (n)

Ans: (A)

Sol: Rest mass of Parent nucleus should be greater than the rest mass of daughter nuclei . Hence Correct option is (A)

Q: The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

(A) 802 nm

(B) 823 nm

(C) 1882 nm

(D) 1648 nm

Ans: (B)

Sol: In hydrogen spectrum series U-V region is Lymen Series .

The Largest wavelength occurs in transition from n=2 to n=1

$\large \frac{1}{122} = R (\frac{1}{1^2} – \frac{1}{2^2} )$ …(i)

The smallest wavelength in infrared region corresponds to

$\large \frac{1}{\lambda} = R (\frac{1}{3^2} – \frac{1}{\infty} )$ …(ii)

On solving (i) & (ii) we get λ = 823 nm

Q: Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is

(A) 1/4

(B) 3/4

(C) 1/2

(D) 1

Ans: (B)

Sol: $\large N = N_0 (\frac{1}{2})^n$

$\large \frac{N}{N_0} = (\frac{1}{2})^2 = \frac{1}{4}$

Probability of Decay $\large = 1- \frac{N}{N_0}$

$\large = 1- \frac{1}{4} = \frac{3}{4}$

Q: If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is  (Mass of the helium nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu)

(A) 7.6 MeV

(B) 56.12 MeV

(C) 10.24 MeV

(D) 23.4 MeV

Ans: (C)

Sol: 4(₂He⁴) = ₈O¹⁶

Mass Defect , Δm = 4 × 4.0026 – 15.9994 = 0.011 amu

The energy released per oxygen nuclei is

= 0.011 × 931.48 MeV

= 10.24 MeV