JEE Advanced 2019 · Paper 2 · Physics Q8

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PhysicsMultiple correct+4 / -1Question 8 of 54

A free hydrogen atom after absorbing a photon of wavelength $\lambda_a$ 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 $\lambda_e$ . Let the change in momentum of atom due to the absorption and the emission are $\Delta p_a$ and $\Delta p_e$ , respectively. If $\lambda_a/\lambda_e = \frac15$ , which of the option(s) is/are correct?

[Use $hc = 1242$ eV nm; 1 nm $= 10^{-9}$ m, $h$ and $c$ are Planck's constant and speed of light, respectively]

m = 2

\lambda_e = 418

\Delta p_a/\Delta p_e = \frac12

The ratio of kinetic energy of the electron in the state

n = m

to the state

n = 1

\frac14

Answer and worked solution

Official answer: A, D

Energies.
$E_a=13.6\left(1-\tfrac1{16}\right)=12.75\text{ eV},\qquad E_e=\frac{E_a}5=2.55\text{ eV}.$

(A) $13.6\left(\frac1{m^2}-\frac1{16}\right)=2.55$ gives $\frac1{m^2}=\frac14$ , so $m=2$ . True.

(B) $\lambda_e=\dfrac{1242}{2.55}=487$ nm, not 418 nm. False.

(C) Each momentum change equals the photon's momentum $h/\lambda$ :
$\frac{\Delta p_a}{\Delta p_e}=\frac{\lambda_e}{\lambda_a}=5.$
False.

(D) The kinetic energy is proportional to $1/n^2$ , so $\dfrac{K_2}{K_1}=\dfrac14$ . True.

(A), (D)

Topics:Bohr model and hydrogen spectrum Photon momentum and radiation pressure

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JEE Advanced 2019 · Paper 2, Physics question 8