Question

Light of wavelength $2000\dot{A}$ falls on a metallic surface whose work function is $4.2\text{eV}$. Calculate the stopping potential. $(hc=12400\text{eV}\dot{A})$

• A. $2\text{V}$
• B. $3\text{V}$
• C. $6\text{V}$
• D. $12\text{V}$

Answer 1

The correct option is A $2\text{V}$

According to Einstein's photoelectric equation,

$K.E_{max}=E-\varphi$

$\Rightarrow e{V}_0=\frac{hc}{\lambda }-\varphi$

$\Rightarrow e{V}_0=\frac{12400\text{eV}\dot{A}}{2000\dot{A}}-4.2\text{eV}$

$\Rightarrow e{V}_0=(6.2-4.2)\text{eV}$

$\Rightarrow e{V}_0=2\text{eV}$

$\Rightarrow V_0=2\text{V}$

Hence, option $\left(B\right)$ is correct.