Question

When radiation of certain wavelength shines on the cathode of the photoelectric cell, the photocurrent produced can be reduced to zero by applying stopping potential of 3 V. If the workfunction of the photo emitter is 3.63 eV, find the frequency of radiation

• A. $3.2\times {10}^{15}$ Hz
• B. $1.6\times {10}^{15}$ Hz
• C. $1.6\times {10}^{12}$ Hz
• D. $3.2\times {10}^{12}$ Hz

Answer 1

The correct option is B $1.6\times {10}^{15}$ Hz

Given,
Stopping potential = $V_s$ = 3V
$\text{Workfunction}=\varphi =3.63eV=3.63\times 1.6\times {10}^{-19}J$

By Einstein photoelectric equation,
$K{E}_{max}=h\nu -\varphi =\frac{hc}{\lambda }-\varphi$

But, $K{E}_{max}=e{V}_s$
$\therefore e{V}_s=h\nu -\varphi$

$\Rightarrow (1.6\times {10}^{-19})\times 3=(6.62\times {10}^{-34})\nu -3.63\times 1.6\times {10}^{-19}$
$\Rightarrow 4.8\times {10}^{-19}=(6.62\times {10}^{-34})\nu -5.808\times {10}^{-19}$
$\Rightarrow 10.608\times {10}^{-19}=(6.62\times {10}^{-34})\nu$
$\Rightarrow \nu =\frac{10.61\times {10}^{-19}}{6.62\times {10}^{-34}}=1.6\times {10}^{15}Hz$