The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A…

Q: The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A° . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

(A) 1215 A°

(B) 1640 A°

(C) 2430 A°

(D) 4687 A°

Ans: (A)

Sol: $\large \frac{\lambda_1}{\lambda_2} = \frac{Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

$\large \lambda_2 = \frac{\lambda_1 Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

On Putting the given values ,

$\large \lambda_2 = \frac{(6561 A^o) ( 1)^2 (\frac{1}{2^2} – \frac{1}{3^2})_1}{(2)^2 (\frac{1}{2^2} – \frac{1}{4^2})_2} $

= 1215 A°

A radioactive sample S1 having an activity of 5 μCi has twice the number of nuclei as another sample S2…

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)

Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Q: Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

(A) $\large \lambda_0 = \frac{2 m c \lambda^2}{h} $

(B) $\large \lambda_0 = \frac{2 h}{m c} $

(C) $\large \lambda_0 = \frac{2 m^2 c^2 \lambda^3}{h^2} $

(D) $\large \lambda_0 = \lambda $

Ans: (A)

Sol: Kinetic Energy of Striking Electrons is

$\large K = \frac{p^2}{2 m} $

As , $\large p = \frac{h}{\lambda}$

$\large K = \frac{h^2}{2 m \lambda^2} $

This is the maximum energy of X-ray photons .

Therefore , $\large \frac{h c}{\lambda_0} = \frac{h^2}{2 m \lambda^2} $

$\large \lambda_0 = \frac{2 m c \lambda^2}{h} $

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

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)

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm…

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