JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Atomic Structure

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm . The change in the velocity (in

cm s^{- 1}

) of He atom after the photon absorption is ____.
(Assume: Momentum is conserved when photon is absorbed.
Use: Planck constant

= 6.6 \times 1 0^{- 34} J s

, Avogadro number

= 6 \times 1 0^{23} mol^{- 1}

, Molar mass of

He = 4 g mol^{- 1}

)

Answer: 30

Solution

The momentum of a photon is

p = \frac{h}{λ}

. When the photon is absorbed, this momentum is transferred to the He atom.

p = \frac{6.6 \times 1 0 ^{- 34}}{330 \times 1 0 ^{- 9}} = 2 \times 1 0^{- 27}

kg·m/s. Mass of one He atom =

\frac{4 \times 1 0 ^{- 3}}{6 \times 1 0 ^{23}} = \frac{2}{3} \times 1 0^{- 26}

kg. Change in velocity

Δ v = \frac{p}{m} = \frac{2 \times 1 0 ^{- 27}}{\frac{2}{3} \times 1 0 ^{- 26}} = \frac{2 \times 1 0 ^{- 27} \times 3}{2 \times 1 0 ^{- 26}} = 3 \times 1 0^{- 1} = 0.3

m/s = 30 cm/s.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Integer answer