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

Answer 1

The correct option is D Graph does not change.

According to Einstein's photoelectric equation,

$K{E}_{max}=hv-{\varphi }_0$

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

$\Rightarrow V_s=\frac{hc}{\lambda e}-\frac{\varphi }{e}...\left(i\right)$

Where $\lambda \to$ wavelength of incident light
${\varphi }_0\to$ work function
$V_s\to$ stopping potential

Comparing equation $\left(i\right)$ with $y=mx+c$, we get slope $=\frac{hc}{e}$

Increasing the frequency of incident radiation has no effect on work function and frequency. So, the graph will not change.

Hence, $\left(D\right)$ is the correct answer.