Question

For photo-electric effect with incident photon wavelength $\lambda$, the stopping potential is $V_0$. Identify the correct variation(s) of $V_0$ with $\lambda$ and $1/\lambda$.

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (C)

Maximum kinetic energy of the photon is given by
$\frac{hc}{\lambda }=e{V}_0+\varphi$
where $\varphi$ is the work function.

On solving further,
$V_0=\frac{hc}{e}\left(\frac{1}{\lambda }\right)-\frac{\varphi }{e}$(which is equation of a straight line )
and the relation between $V_0$ and $\lambda$ is
$V_0\propto \frac{1}{\lambda }$
which shows straight line graph. Hence option $C$ is correct.

Now the relation between stopping potential $\left(V_0\right)$ and $\lambda$.
From the equation, $\frac{hc}{\lambda }=e{V}_0+\varphi$
When, $\lambda \to 0,V_0\to \infty \lambda \to {\lambda }_{threshold},V_0\to 0$

So, option A is correct.