Question

The graph between the stopping potential $\left(V_0\right)$ and wave number $\left(\frac{1}{\lambda }\right)$ is as shown in the figure. $\varphi$ is the work function, then

• A. ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
• B. ${\varphi }_1:{\varphi }_2:{\varphi }_3=4:2:1$
• C. $\text{tan}\theta \propto \text{hc/e}$ where $\tan \theta$ is the slope
• 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 $\text{tan}\theta \propto \text{hc/e}$ where $\tan \theta$ is the slope

$e{V}_0=K{E}_{max}=\frac{hc}{\lambda }-\varphi$
$V_0=\left(\frac{hc}{e}\right)\frac{1}{\lambda }-\frac{\varphi }{e}$
$\Rightarrow$ for $V_0=0$
$\frac{hc}{{\lambda }_1}={\varphi }_1$
$\frac{hc}{{\lambda }_2}={\varphi }_2$
$\frac{hc}{{\lambda }_3}={\varphi }_3$
From the graph
$\frac{1}{{\lambda }_1}=0.001$
$\frac{1}{{\lambda }_2}=0.002$
$\frac{1}{{\lambda }_3}=0.004$
Hence
${\varphi }_1=0.001\times hc$
${\varphi }_2=0.002\times hc$
${\varphi }_3=0.004\times hc$
${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$