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. $tan\theta \propto \left(hc\right)/e$
• 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 $tan\theta \propto \left(hc\right)/e$

The work function is given by:
$\varphi =\frac{hc}{{\lambda }_0}$ where $\lambda$ is the threshold wavelength.
Therefore,
${\varphi }_1=0.001\times hc$
${\varphi }_2=0.002\times hc$
${\varphi }_3=0.004\times hc$
Hence, ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$ option A is correct
Now, stopping potential is given by
$V_0=\frac{K}{e}=\frac{1}{e}(\frac{hc}{\lambda }-\varphi )$
i.e. $V_0=\frac{hc}{e}\frac{1}{\lambda }-\frac{\varphi }{e}$
This is the equation of the straight line defining the graph between $V_0$ and $\frac{1}{\lambda }$ with slope i.e. $\tan \theta =\frac{hc}{e}$. Thus, option C is correct
The UV light wavelength vary from 10nm to 380nm.
Here the cutoff wavelengths are 1000nm,500nm and 250nm. Thus UV light can be used to emit photoelectrons from any of these metals. Thus, option D is incorrect.