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)

$\frac{h c}{\lambda}-\phi =eV_{0}$
$v_{0} =\frac{h c}{e \lambda}-\frac{\phi}{e}$
For metal A
$\frac{\phi A}{h c}=\frac{1}{\lambda}$
For metal B
$\frac{\phi B}{h c}=\frac{1}{\lambda}$
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 $\left(\phi\right).$