Question

Work function of caesium is $2.14eV$. What is the wavelength of incident light if photocurrent is brought to zero by a stopping potential of $0.6V$?

• A. $454nm$
• B. $113nm$
• C. $906nm$
• D. $226nm$

Answer 1

The correct option is A $454nm$

Einstein's equation : $E=\varphi +K.E_{max}$
Given, work function $\varphi =2.14eV$
and we know $E=\frac{hc}{\lambda }$
$\therefore \frac{hc}{\lambda }=2.14+K.E_{max}$

For stopping potential $V_0$, $K.E_{max}=e{V}_0=0.6eV$
$\Rightarrow \frac{hc}{\lambda }=2.14eV+0.6eV=2.74eV$
$\Rightarrow \frac{12431}{\lambda }=2.74eV$
$\Rightarrow \lambda =\frac{12431}{2.74}=4536.86\stackrel{\circ }{A}$
$=453.68nm\approx 454nm$