Given below are two statements:
**Statement I:** The figure shows the variation of stopping potential with frequency $(\nu)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\dfrac{h}{e}$, where $h$ is Planck's constant and $e$ is the charge of an electron.
**Statement II:** $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation of the same frequency.
In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (D) Statement I is correct and Statement II is incorrect
$eV_0 = h\nu - \phi \Rightarrow V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}$. The slope is $\dfrac{h}{e}$, so Statement I is correct.
$M_2$ has the larger threshold frequency, so a larger work function. For the same frequency its photoelectrons have **less** kinetic energy. Statement II is incorrect.
Solution by Sreeraj P, M.Sc Physics