Question

The graph between the stopping potential $\left(V_0\right)and\left(1/\lambda \right)$ is shown in the Fig. ${\varphi }_1,{\varphi }_{2}and{\varphi }_3$ are work functions. Then, which of the following is/are correct?

• A. ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
• B. ${\varphi }_1:{\varphi }_2:{\varphi }_3=4:2:1$
• C. tan $\theta \alpha$ (hc)/e
• D. Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

Answer 1

Answer: (A), (C)

The correct options are
A ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
C tan $\theta \alpha$ (hc)/e

${\varphi }_1:{\varphi }_2:{\varphi }_3=e{V}_{0_1}:e{V}_{0_2}:e{V}_{0_3}$

$V_{0_1}:V_{0_2}:V_{0_3}=0.001:0.002:0.004=1:2:4$
Therefore, option(a) is correct.
By Einstein's photoelectric equation, $\frac{hc}{\lambda }-\varphi =eV$
$\Rightarrow V=\frac{hc}{e\lambda }-\frac{\varphi }{e}$ (i)
Comparing Eq. (i) by y = mx + c, we get the slope of the line m
$\frac{hc}{e}=tan\theta$
$\Rightarrow$ Option (c) is correct.
From the graph it is clear that,
$\frac{1}{{\lambda }_{0_1}}=0.001n{m}^{-1}$
$\Rightarrow {\lambda }_{0_1}=\frac{1}{0.001}=1000nm$
Also, $\frac{1}{{\lambda }_{0_2}}=0.002n{m}^{-1}\Rightarrow {\lambda }_{0_2}=500nmand{\lambda }_{0_2}=250nm$
Violet color light will have wavelength less than 400 nm.
Therefore, this light will be unable to show the photoelectric effect on plate 3 $\Rightarrow$ Option (d) is wrong.