Question

The vapours of $Hg$ absorb some electrons accelerated by a potential difference of $4.5$ volt, as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by single $Hg$ atom, then the wave number$\left(1/\lambda \right)$ of the light is:

• A. $3.63\times {10}^{6}metr{e}^{-1}$
• B. $400\times {10}^{6}metr{e}^{-1}$
• C. $36.3\times {10}^{6}metr{e}^{-1}$
• D. $22\times {10}^{6}metr{e}^{-1}$

Answer 1

The correct option is A $3.63\times {10}^{6}metr{e}^{-1}$

Energy of an accelerated electron $=qV$, where $q$ is charge of electron and $V$ is the potential difference.
$qV=1.602\times {10}^{-19}\times 4.5$$=7.209\times {10}^{-19}J$
The electron energy is completely converted to light.
$\therefore \frac{hc}{\lambda }=7.209\times {10}^{-19}$
$\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{\lambda }=7.209\times {10}^{-19}$
$\therefore$ Wave number $\overline{v}=\left(1/\lambda \right)=3.63\times {10}^6$ $metr{e}^{-1}$.