Question

The vapours of $Hg$ absorb some electrons accelerated by a potential difference of $4.5\text{eV}$, 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 a single $Hg$ atom, find the wave number $\frac{1}{\lambda }$ of the light.

(Take: Planck's constant (h) =$6.6\times {10}^{-34}Js$)

• A. $7.26\times {10}^8{\text{m}}^{-1}$
• B. $8.32\times {10}^{10}{\text{m}}^{-1}$
• C. $3.63\times {10}^6{\text{m}}^{-1}$
• D. $3.08\times {10}^{10}{\text{m}}^{-1}$

Answer 1

The correct option is C $3.63\times {10}^6{\text{m}}^{-1}$

Energy of an accelerated electron
$=1.6\times {10}^{-19}\times 4.5=7.2\times {10}^{-19}\text{J}$

This energy is completely converted into light.
We know that,
$\frac{hc}{\lambda }=7.2\times {10}^{-19}$
$\Rightarrow \frac{6.6\times {10}^{-34}\times 3\times {10}^8}{\lambda }=7.2\times {10}^{-19}$
$\Rightarrow \frac{1}{\lambda }=\frac{7.2\times {10}^{-19}}{6.6\times {10}^{-34}\times 3\times {10}^8}$
$\Rightarrow \frac{1}{\lambda }=\overline{\nu }=3.63\times {10}^6{\text{m}}^{-1}$