Question

The graph $\frac{1}{\lambda }$ and stopping potential $\left(V\right)$ of three metals having work function ${\varphi }_1$, ${\varphi }_2$ and ${\varphi }_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is/are correct? [Here $\lambda$ is the wavelength of the incident ray]
(i) Ratio of work functions ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
(ii) Ratio of work functions ${\varphi }_1:{\varphi }_2:{\varphi }_3=4:2:1$
(iii) $\tan \theta \propto \frac{hc}{e}$, where $h=$ Planck's constant, $c=$ speed of light
(iv) The violet colour-light can eject photoelectrons from metals 2 and 3.

• A. (i), (iii)
• B. (i), (iv)
• C. (ii), (iii)
• D. (i), (ii) and (iv)

Answer 1

Answer: (A)

(i), (iii)

According to Einstein photoelectric equation,
$E=K_{max}+\varphi$
$\frac{hc}{\lambda }=\varphi +e{V}_S$
$\therefore V_S=[\frac{hc}{\lambda }-\varphi ]\frac{1}{e}$
$V_S$ versus $\frac{1}{\lambda }$ graph is a straight line.
Slope, $\tan \theta =\frac{hc}{e}$
The curves intersect the $\frac{1}{\lambda }$ axis, where $V_S$ is zero.
$\therefore \frac{hc}{\lambda }-\varphi =0$
$\Rightarrow \frac{1}{\lambda }=\frac{\varphi }{hc}$

For metal 1	For metal 2	For metal 3
$\frac{{\varphi }_1}{hc}=0.001$	$\frac{{\varphi }_2}{hc}=0.002$	$\frac{{\varphi }_3}{hc}=0.004$

$\therefore {\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
For metal 2, threshold wavelength,
$\lambda o_2=\frac{hc}{\varphi }=\frac{1}{0.002}=500nm$
Similarly, $\lambda o_3=250nm$
${\lambda }_{violet}=400nm$ So ${\lambda }_{violet}<{\lambda }_{threshold}$ for metal 1, ${\lambda }_{violet}>{\lambda }_{threshold}$ for metal 3.
Thus, violet will eject photoelectron for metal 2 and not metal 3.