Question

The threshold frequency of a certain material is $3\times {10}^{14}\text{Hz}.$ If it is illuminated by a radiation of frequency $6\times {10}^{14}\text{Hz}$, then the stopping potential will be

$[\text{Take}h=6.62\times {10}^{-34}\text{Js}]$

• A. $1.98\text{V}$
• B. $2.54\text{V}$
• C. $0.65\text{V}$
• D. $1.24\text{V}$

Answer 1

The correct option is D $1.24\text{V}$

Given:
${\nu }_0=3\times 1^{14}\text{Hz}$
$\nu =6\times {10}^{14}\text{Hz}$

According to Einstein's photoelectric eqution,

$e{V}_0=E-\varphi$

$\Rightarrow e{V}_0=h\nu -h{\nu }_0$

$\Rightarrow V_0=\frac{h(\nu -{\nu }_0)}{e}$

$\Rightarrow V_0=\frac{6.62\times {10}^{-34}\times (6\times {10}^{14}-3\times {10}^{14})}{1.6\times {10}^{-19}}$

$\Rightarrow V_0=1.24\text{V}$

Hence, $\left(D\right)$ is the correct answer.