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ChemistryInteger+4Question 37 of 57

JEE Advanced 2021 · Paper 2, Chemistry question 18

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s−1^{-1}) of He atom after the photon absorption is ____.

(Assume: Momentum is conserved when photon is absorbed.

Use: Planck constant =6.6×10−34= 6.6 \times 10^{-34} J s, Avogadro number =6×1023= 6 \times 10^{23} mol−1^{-1}, Molar mass of He =4= 4 g mol−1^{-1})

Answer and worked solution

Official answer: 30

Momentum conservation. The atom gains the photon's momentum hλ\frac h\lambda, so Δv=hmλ\Delta v=\dfrac h{m\lambda}.

Numbers.

  • Mass of one He atom: m=4×10−36×1023=6.67×10−27m=\dfrac{4\times10^{-3}}{6\times10^{23}}=6.67\times10^{-27} kg.
  • Photon momentum: hλ=6.6×10−34330×10−9=2×10−27\dfrac h\lambda=\dfrac{6.6\times10^{-34}}{330\times10^{-9}}=2\times10^{-27} kg m s−1^{-1}.

Velocity change.
Δv=2×10−276.67×10−27=0.30 m s−1=30 cm s−1.\Delta v=\frac{2\times10^{-27}}{6.67\times10^{-27}}=0.30\text{ m s}^{-1}=30\text{ cm s}^{-1}.

Topics:de Broglie and HeisenbergPhotoelectric effect and spectra
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