Question

Consider a helium $(He)$ atom that absorbs a photon of wavelength $330nm$. 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 $=6.6\times 10^{-34}Js$, Avogadro number $=6\times 10^{23}mo{l}^{-1}$,
Molar mass of $He=4gmo{l}^{-1}$ )

Answer 1

Answer: 30

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