In an experiment on photoelectric effect, a student plots stopping potential $\mathrm{V}_0$ against…

In an experiment on photoelectric effect, a student plots stopping potential $\mathrm{V}_0$ against reciprocal of the wavelength $\lambda$ of the incident light for two different metals $\mathrm{A}$ and $\mathrm{B}$. These are shown in the figure.
Looking at the graphs, you can most appropriately say that:

1. Work function of metal $\mathrm{B}$ is greater than that of metal $\mathrm{A}$
2. For light of certain wavelength falling on both metal, maximum kinetic energy of electrons emitted from A will be greater than those emitted from $\mathrm{B}$.
3. Work function of metal $\mathrm{A}$ is greater than that of metal $\mathrm{B}$
4. Students data is not correct

Solution

$\begin{aligned} & \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_0 \\ & \mathrm{v}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi}{\mathrm{e}} \end{aligned}$
As the value of $\frac{1}{\lambda}$ (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function $(\phi)$.

Asked in: JEE Main 2013 (25 Apr Online)

Practice more Dual Nature of Matter and Radiation questions on Aicharya