Question

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 $He =4\, g\, mol ^{-1}$ )

Answer 1

$\lambda=\frac{ h }{ m (\Delta V )}$
$330 \times 10^{-9}=\frac{6.6 \times 10^{-34}}{\left(\frac{4 \times 10^{-3}}{6 \times 10^{23}}\right) \times \Delta V }$
$\Delta V =\frac{6.6 \times 6 \times 10^{23} \times 10^{-34}}{4 \times 10^{-3} \times 330 \times 10^{-9}}$
$=0.30 \,m / s$
$=0.30 \times 100\, cm / s$
$=30 \,cm / s$