Question

If a photocell is illuminated with a radiation of $1240{\text{A}}^{\circ }$, then stopping potential is found to be $8\text{V}$ . The work function of the emitter and the threshold wavelength are

• A. $1\text{eV},5200\stackrel{\circ }{\text{A}}$
• B. $2\text{eV},6200\stackrel{\circ }{\text{A}}$
• C. $3\text{eV},7200\stackrel{\circ }{\text{A}}$
• D. $4\text{eV},4200\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is B $2\text{eV},6200\stackrel{\circ }{\text{A}}$

The Einstein's equation for photoelectric effect
$K{E}_{max}=\frac{hc}{\lambda }-W$
(where $W$ = work function of metal,$\lambda =$ wavelength of incident light)
If $V_0$ be the stopping potential, then $K{E}_{max}=e{V}_0$
so, ${\text{eV}}_0=\frac{hc}{\lambda }-W$
$W=\frac{hc}{\lambda }-e{V}_0=\frac{(6.62\times {10}^{-34})\times (3\times {10}^8)}{1240\times {10}^{-10}}-(1.6\times {10}^{-19})\left(8\right)$
$=3.2\times {10}^{-19}$
$=2eV(\because 1eV=1.6\times {10}^{-19}J)$
If ${\lambda }_0$ be the threshold wavelength, $W=\frac{hc}{{\lambda }_0}$
${\lambda }_0=\frac{hc}{W}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3.2\times {10}^{-19}}$
$\Rightarrow {\lambda }_0=6200\stackrel{\circ }{\text{A}}$