Question

The work function of a surface of a photosensitive material is $6.2eV$. The wavelength of the incident radiation for which the stopping potential is $5V$ lies in the:

• A. Infrared region
• B. X-ray region
• C. Ultraviolet region
• D. Visible region

Answer 1

The correct option is C

Ultraviolet region

Explanation for the correct option

Step 1: Given

Work function is $6.2eV$

Stopping potential $V=5V$

Step 2: Formula used

We have to use work function formula . Total energy of photon hitting a metal surface is equal to sum of work function of metal and energy of ejected electron.

$\frac{hc}{\lambda }=W+eV$, where $h$ is Planck's constant$=6.62\times {10}^{-34}Js$, $W=$work function, $c=$speed of light$=3\times {10}^{8}m{s}^{-1}$, $\lambda =$wavelength, $1eV=1.6\times {10}^{-19}C$

Step 3: Solution for wavelenth

$$\begin{gathered} \frac{hc}{\lambda }=6.2eV+5eV \\ \lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{11.2\times 1.6\times {10}^{-19}} \\ \lambda =1.108\times {10}^{-7} \\ \lambda =110.8nm \end{gathered}$$

This wavelength lies in the ultraviolet region

Explanation for the incorrect option

Option (A)

The wavelength in the infrared region ranges from $700nm-1000nm$, since the wavelength is only $110.8nm$ so it is not in the infrared region.

Option (B)

The wavelength in the X-ray region ranges from $0.01nm-10nm$, since the wavelength is only $110.8nm$ so it is not in the X-ray region .

Option (D)

The wavelength in the Visible region ranges from $380nm-700nm$, since the wavelength is only $110.8nm$ so it is not in the Visible region

Hence, option C is correct.