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 $V=4.5eV$

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

$\therefore E=\frac{hc}{\lambda }=4.5eV$

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

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