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 × 10 -34 J s , Avogadro number =6 × 10 23 mol -1 , Molar mass of He =4 g mol -1 )

Chemistry · Structure of Atom

JEE Advanced 2021 — Paper 2 — Question 16

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm . The change in the velocity (in $\mathrm{cm} \mathrm{s}^{-1}$ ) of He atom after the photon absorption is $\qquad$ . (Assume: Momentum is conserved when photon is absorbed. Use: Planck constant $=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ , Avogadro number $=6 \times 10^{23} \mathrm{~mol}^{-1}$ , Molar mass of $\mathrm{He}=4 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

$\quad 30$ $\lambda=\frac{\mathrm{h}}{\mathrm{m}(\Delta \mathrm{V})}$

$330 \times 10^{-9}=\frac{6.6 \times 10^{-34}}{\left(\frac{4 \times 10^{-3}}{6 \times 10^{23}}\right) \times \Delta V}$

$\Delta \mathrm{V}=\frac{6.6 \times 6 \times 10^{23} \times 10^{-34}}{4 \times 10^{-3} \times 330 \times 10^{-9}}$

$=0.30 \mathrm{~m} / \mathrm{s}$

$=0.30 \times 100 \mathrm{~cm} / \mathrm{s}$

$=30 \mathrm{~cm} / \mathrm{s}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Chemistry
Chapter: Structure of Atom
Topic: Wave-Particle Duality of Matter - de Broglie, Heisenberg