Question

The vapours of Hg absorb some electrons accelerated by a potential difference of 4.5 volts 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, then the wave number of the light is:

• A. $3.63\times {10}^6{m}^{-1}$
• B. $3.88\times {10}^6{m}^{-1}$
• C. $3.98\times {10}^6{m}^{-1}$
• D. $3.56\times {10}^6{m}^{-1}$

Answer 1

Answer: (A)

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

Wave number is given by $\overline{\nu }=\frac{1}{\lambda }$
If the full energy of a single incident electron is supposed to be converted into light emitted by a single Hg atom then,
$\frac{hc}{\lambda }=4.5eV$
$\overline{\nu }=\frac{1}{\lambda }=\frac{4.5eV}{hc}=\frac{4.5eV}{1242eV-nm}=3.63\times {10}^6{m}^{-1}$