Question

In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

• A. Straight line shifts to right
• B. Straight line shifts to left
• C. Slope of the straight line get more steep
• D. Graph does not change

Answer 1

The correct option is D

Graph does not change

The explanation for correct option:

Option (d)

Step 1: Understand the photoelectric experiment

1. The photoelectric effect occurs when electromagnetic radiation, such as light, strikes a substance and emits electrons.
2. Photoelectrons are electrons emitted in this manner.
3. From the photoelectric experiment:

$\mathrm{\varphi }={\mathrm{\varphi }}_0+\mathrm{eV}\Rightarrow \frac{\mathrm{hc}}{\mathrm{\lambda }}={\mathrm{\varphi }}_0+\mathrm{eV}$

$\mathrm{eV}=\frac{\mathrm{hc}}{\mathrm{\lambda }}-{\mathrm{\varphi }}_0\Rightarrow \mathrm{V}=\frac{\mathrm{hc}}{\mathrm{\lambda e}}-\frac{{\mathrm{\varphi }}_0}{\mathrm{e}}$

$\mathrm{V}=\left(\frac{\mathrm{hc}}{\mathrm{e}}\right)\left(\frac{1}{\mathrm{\lambda }}\right)-\frac{{\mathrm{\varphi }}_0}{\mathrm{e}}$

Step 2: Comparison with the equation of a straight line.

1. On comparing the above equation with the equation of a straight line, i.e., $\mathrm{y}=\mathrm{m}\mathrm{x}+\mathrm{c}$, we get
   Slope: $\mathrm{m}=\frac{\mathrm{hc}}{\mathrm{e}}$ ,
2. $\mathrm{y}$-intercept $\mathrm{c}=-\frac{{\mathrm{\varphi }}_0}{\mathrm{e}}$
   

Step 3: Analyze the graph

1. Now, if we change the intensity of incident radiation, then $\mathrm{h},\mathrm{c},\mathrm{e}$ and ${\mathrm{\varphi }}_0$ will not change.
2. Therefore, neither the slope of the graph nor the y-intercept cut by the graph will change on changing the intensity of incident radiation.
3. So, the graph will not change.

Hence, option (d) is correct.