JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Modern Physics

A free hydrogen atom after absorbing a photon of wavelength

λ_{a}

gets excited from the state

n = 1

to the state

n = 4

. Immediately after that the electron jumps to

n = m

state by emitting a photon of wavelength

λ_{e}

. Let the change in momentum of atom due to the absorption and the emission are

Δ p_{a}

and

Δ p_{e}

, respectively. If

λ_{a} / λ_{e} = \frac{1}{5}

. Which of the option(s) is/are correct ?
[Use

hc = 1242 eVnm; 1 nm = 1 0^{- 9} m

, h and c are Planck's constant and speed of light, respectively]

(A) $λ_{e} = 418 nm$
(B)The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$ ✓ Correct
(C)m = 2✓ Correct
(D) $Δ p_{a} /Δ p_{e} = \frac{1}{2}$

Answer: (B), (C)

Solution

Energy absorbed:

\frac{h c}{λ _{a}} = 13.6 [1 - \frac{1}{16}] = 13.6 \times \frac{15}{16}

Energy emitted:

\frac{h c}{λ _{e}} = 13.6 [\frac{1}{m ^{2}} - \frac{1}{16}]

Taking ratio with

\frac{λ _{a}}{λ _{e}} = \frac{1}{5}

\frac{1}{m ^{2}} - \frac{1}{16} = \frac{15}{16} \times \frac{1}{5} = \frac{3}{16}

Solving:

m = 2

. Option (C) is correct.

Calculating

λ_{e}

λ_{e} = \frac{1242 \times 16}{13.6 \times 3} nm \approx 486 nm

This doesn't match option (A), which is incorrect.

Kinetic energy ratio:

\frac{K E _{2}}{K E _{1}} = \frac{1}{2 ^{2}} = \frac{1}{4}

. Option (B) is correct.

Momentum ratio:

\frac{Δ p _{a}}{Δ p _{e}} = \frac{λ _{e}}{λ _{a}} = 5

, not

\frac{1}{2}

. Option (D) is incorrect.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options