Question

Electromagnetic radiation having $\mathrm{\lambda }=310\mathrm{A}$ is subjected to a metal sheet having a work function = $12.8\mathrm{eV}$. What will be the velocity of photoelectrons having maximum kinetic energy?

Answer 1

Step 1: Given information

Wavelength of light,

$$\begin{gathered} \mathrm{\lambda }=310\mathrm{A} \\ =310\times {10}^{-10} \end{gathered}$$

Work function = $12.8\mathrm{eV}$

Step 2: To find

We have to find the velocity of photoelectrons having maximum kinetic energy.

Step 3: Finding the work function

We know the formula,

$\mathrm{E}\left(\mathrm{photon}\right)=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

Here,

h is Planck's constant.

c is the speed of light.

By substituting the values,

$$\begin{gathered} =\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{(31\times {10}^{-9})} \\ =6.406\times {10}^{-18}\mathrm{J} \end{gathered}$$

Work function is,

$$\begin{gathered} \mathrm{W}=12.8\mathrm{eV} \\ =12.6\times 1.6\times {10}^{-19} \\ =2.016\times {10}^{-18}\mathrm{J} \end{gathered}$$

Step 4: Finding the maximum kinetic energy of the photoelectron

We know that,

$$\begin{gathered} (0.5){{\mathrm{mv}}^2}_{\max }=\mathrm{E}\left(\mathrm{photon}\right)-\mathrm{W} \\ =6.406\times {10}^{-18}-2.016\times {10}^{-18} \\ =4.39\times {10}^{-18}\mathrm{J} \end{gathered}$$

The velocity will be:

$$\begin{gathered} {{\mathrm{v}}^2}_{\max }=\frac{(4.39\times {10}^{-18})\times 2}{(9.1\times {10}^{-31})} \\ =9.648\times {10}^{12} \\ {\mathrm{v}}_{\max }=\sqrt{9.648\times {10}^{12}} \\ =3.106\times {10}^6{\mathrm{ms}}^{-1} \end{gathered}$$

Therefore, the velocity of photoelectrons having maximum kinetic energy is $3.106\times {10}^6{\mathrm{ms}}^{-1}$.