Question

The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest 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 electron in a single Hg atom, find the wave number $\left(\frac{1}{\lambda }\right)$ of the light

• A. $3.63\times 10^{6} \, \text{m}^{- 1}$
• B. $5.93\times 10^{6} \, \text{m}^{- 1}$
• C. $5.93\times 10^{6} \, \text{c}\text{m}^{- 1}$
• D. $5.62\times 10^{6} \, \text{m}^{- 1}$

Answer 1

Answer: (A)

Potential difference $V = 4.5 \, eV$

$\therefore \, \, \, $ Potential energy $=4.5 \, eV$ energy absorbed and this energy is emitted by electron of Hg atom.

$\therefore \, \, E=\frac{h c}{\lambda } \, =4.5 \, eV$

$=4.5\times 1.6\times 10^{- 19} \, \text{J}$

$\therefore \, \frac{1}{\lambda }=\frac{4.5 \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34} \times 3.0 \times 10^{8}} \, \text{m}^{- 1} \, \approx 3.63\times 10^{6} \, \text{m}^{- 1}$