Question

A radiation of wavelength $3000A^{\circ }$ falls on a photoelectric surface of work function 1.6 eV. What is the stopping potential of the emitted electron?

• A. 3.52 V
• B. 2.53 V
• C. 1.50 V
• D. 3.32 V

Answer 1

The correct option is B 2.53 V

Given,
$\text{Wavelength of radiation}\left(\lambda \right)=3000A^{\circ }=3000\times {10}^{-10}m$
$\text{Work function}\left(\varphi \right)=1.6eV=1.6\times 1.6\times {10}^{-19}J=2.56\times {10}^{-19}J$

According to Einstein's Photoelectric equation,
$K{E}_{max}=h\nu -\varphi =\frac{hc}{\lambda }-\varphi$

We know,
$K{E}_{max}=e{V}_s$,
Where, $V_s$ is the stopping potential.
Hence,
$e{V}_s=\frac{hc}{\lambda }-\varphi$

$\therefore (1.6\times {10}^{-19})\times V_s=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{3000\times {10}^{-10}}-2.56\times {10}^{-19}$

$\Rightarrow (1.6\times {10}^{-19})\times V_s=6.62\times {10}^{-19}-2.56\times {10}^{-19}$

$\Rightarrow V_s=\frac{4.06\times {10}^{-19}}{1.6\times {10}^{-19}}=2.53V$