Question

The minimum energy required to remove the electron from metal surface is $7.50\times {10}^{-19}J$. What will be the stopping potential required when this metal surface is exposed to $U.V$. light of $\lambda =375\mathring{A}$?

Answer 1

Photoelectric equation is given by,

$K.E_{max}=\frac{hc}{\lambda }-W$

Where $W$ is the work function of the metal i.e minimum energy required to eject an electron and $\lambda$ is the wavelength of the incident light.

Energy of incident light is given by $E=\frac{hc}{\lambda }$

Substituting $h,c$ and $\lambda =375A^0=3.75\times {10}^{-8}m$

$E=\frac{(6.63\times {10}^{-34}Js)(3\times {10}^{8}m/s)}{3.75\times {10}^{-8}m}$

$E=5.3\times {10}^{-18}J$

$K.E_{max}=E-W=45.5\times {10}^{-19}J$

$K.E_{max}=e{V}_0$ where $V_0$ is the stopping potential

$45.5\times {10}^{-19}J=e{V}_0$

$V_0=28.4V$