Question

The electric field of light wave is given as $\overrightarrow{E}={10}^3\cos (\frac{2\pi x}{5\times {10}^{-7}}-2\pi \times 6\times {10}^{14}t)\hat{x}\frac{N}{C}$
This light falls on a metal plate of work function $2\text{eV}$. The stopping potential of the photo-electrons is

Given, $E\left(\text{in}\text{eV}\right)=\frac{12375}{\lambda \left(\text{in}\AA \right)}$

• A. $2.48\text{V}$
• B. $0.72\text{V}$
• C. $0.48\text{V}$
• D. $2.0\text{V}$

Answer 1

The correct option is C $0.48\text{V}$

Given:
$\omega =2\pi \times 6\times {10}^{14}$
$\Rightarrow f=6\times {10}^{14}\text{Hz}$

Wavelength
$\lambda =\frac{c}{f}=\frac{3\times {10}^8}{6\times {10}^{14}}=0.5\times {10}^{-6}\text{m}=5000\AA$

$E=\frac{12375}{\lambda \left(\text{in}\AA \right)}$

Now, $E=\frac{12375}{5000}=2.48\text{eV}$

Given that $W=2\text{eV}$

Using $E=W+e{V}_s$

$\Rightarrow 2.48\text{eV}=2\text{eV}+\text{e}{V}_s$

$\Rightarrow V_s=0.48\text{V}$

Hence option $\left(C\right)$ is correct.