Use Einstein’s photoelectric equation

For a photosensitive material, $K_{\max} = h\nu - \phi$ and the stopping potential $V_s$ is given by $eV_s = K_{\max} = h\nu - \phi$

So, $V_s = \frac{h}{e}\nu - \frac{\phi}{e}$

This is a straight-line relation between $V_s$ and frequency $\nu$.

Analyse Statement I

From $V_s = \frac{h}{e}\nu - \frac{\phi}{e}$ the slope of the graph of stopping potential versus frequency is $\frac{dV_s}{d\nu} = \frac{h}{e}$

This slope is independent of the material.

Therefore, Statement I is correct.

Analyse Statement II

For the same incident frequency, $K_{\max} = h\nu - \phi$

The material with smaller work function $\phi$ emits photoelectrons with greater maximum kinetic energy.

On the $V_s$ vs $\nu$ graph, the line with smaller threshold frequency $\nu_0$ corresponds to smaller work function because $\phi = h\nu_0$

From the given figure (as implied in the question), $M_1$ has the lower threshold frequency, so $M_1$ has smaller work function and hence gives greater kinetic energy for the same frequency.

Therefore, the statement that $M_2$ will emit photoelectrons of greater kinetic energy for the same frequency is incorrect.

Hence, Statement II is incorrect.

Conclusion

• Statement I: Correct
• Statement II: Incorrect

Therefore, the correct option is: $\boxed{\text{A}}$