Question

In an experiment on photoelectric effect, a student plots stopping potential $V_0$ against reciprocal of the wavelength $\lambda$ of the incident light for two different metals $A$ and $B$. These are shown in the figure.

Looking at the graphs, you can most appropriately say that:

• A. Work function of metal B is greater than that of metal A
• B. For light of certain wavelength falling on both metal, maximum kinetic energy of electrons emitted from A will be greater than those emitted from B
• C. Work function of metal A is greater than that of metal B
• D. Students data is not correct

Answer 1

Answer: (D)

$\frac{hc}{\lambda }-\varphi =e{V}_0$
$v_0=\frac{hc}{e\lambda }-\frac{\varphi }{e}$
For metal A
$\frac{\varphi A}{hc}=\frac{1}{\lambda }$
For metal B
$\frac{\varphi B}{hc}=\frac{1}{\lambda }$
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 $\left(\varphi \right).$