Photoelectric equation

For incident light of wavelength $\lambda$, the maximum kinetic energy of emitted electrons is $K_{\max} = h\nu - \phi$ where $\phi$ is the work function.

Since $\nu = \frac{c}{\lambda},$ we get $K_{\max} = \frac{hc}{\lambda} - \phi.$

Relation with stopping potential

The stopping potential $V_s$ is related to maximum kinetic energy by $eV_s = K_{\max}.$

Therefore, $eV_s = \frac{hc}{\lambda} - \phi$ or $V_s = \frac{hc}{e}\left(\frac{1}{\lambda}\right) - \frac{\phi}{e}.$

Interpretation of the graph

This is a straight line in $V_s$ versus $\dfrac{1}{\lambda}$: $V_s = m\left(\frac{1}{\lambda}\right) + c$ with $m = \frac{hc}{e}, \qquad c = -\frac{\phi}{e}.$

So:

• Slope depends only on $h, c, e$
• Intercept depends only on work function $\phi$

Effect of increasing intensity

Increasing the intensity of incident radiation:

• increases the number of emitted photoelectrons
• does not change their maximum kinetic energy
• hence does not change stopping potential for a given wavelength

Therefore, the graph of $V_s$ versus $1/\lambda$ remains unchanged.

Checking options

• A: Slope becomes steeper $\rightarrow$ Incorrect, slope is $\dfrac{hc}{e}$ and is independent of intensity.
• B: Graph does not change $\rightarrow$ Correct.
• C: Straight line shifts left $\rightarrow$ Incorrect.
• D: Straight line shifts right $\rightarrow$ Incorrect.

Final answer

$\boxed{\text{B: Graph does not change}}$