JEE Advanced 2021 Paper 2 · Q18 · 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 = $6.6 \times 10^{-34}$ J s, Avogadro number = $6 \times 10^{23}$ mol$^{-1}$, Molar mass of He = 4 g mol$^{-1}$.)
Reveal answer + step-by-step solution
Correct answer:30
Solution
Momentum of photon $p = h/\lambda$. By conservation, the He atom gains this momentum: $m \Delta v = h/\lambda$ $\Rightarrow$ $\Delta v = h/(m\lambda)$.
Mass of one He atom $m = (4 \times 10^{-3})/(6 \times 10^{23}) = 6.667 \times 10^{-27}$ kg.
$\Delta v = (6.6 \times 10^{-34})/[(6.667 \times 10^{-27})(330 \times 10^{-9})] = (6.6 \times 10^{-34})/(2.2 \times 10^{-33}) = 0.30$ m s$^{-1}$ $= 30$ cm s$^{-1}$.
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