Question

The vapours of $Hg$ absorb some electrons accelerated by a potential diff. of $4.5volt$ as a result of which light is emitted. If the full energy of single incident $e^{-}$ is supposed to be converted into light emitted by single $Hg$ atom, find the wave no. (in $m^{-1}$) of the light.

Answer 1

Energy of an accelerated electron $=S.V=1.602\times {10}^{-19}\times 4.5$

$=7.204\times {10}^{-19}J$

Given that the energy is completed & converted into light

$\frac{bc}{\lambda }=7.209\times {10}^{-9}$

$\therefore \frac{6.625\times {10}^{-34}\times 3\times {10}^8}{\lambda }=7.209\times {10}^{-9}$

$\frac{1}{\lambda }=3.63\times {10}^6{m}^{-1}$