Question

Given below are two statements: Statement I: Figure shows the variation of stopping potential with frequency ($\nu$) for the two photosensitive materials $M_1$ and $M_2$. The slope gives value of $\frac{h}{e}$, where h is Planck's constant, e is the charge of electron. Statement II: $M_2$ will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements, choose the most appropriate answer from the options given below.

• A. Statement I is correct and Statement II is correct
• B. Statement I is correct and Statement II is incorrect
• C. Statement I is incorrect and Statement II is correct
• D. Statement I is incorrect and Statement II is incorrect

Answer 1

Answer: (B)

Statement I is correct and Statement II is incorrect

Answer 2

Statement I: The photoelectric effect equation is given by $h\nu = \phi + K_{max}$, where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, $\phi$ is the work function, and $K_{max}$ is the maximum kinetic energy. The stopping potential $V_0$ is related by $K_{max} = eV_0$. Thus, $h\nu = \phi + eV_0$. Rearranging, $V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$. This is a linear equation in $V_0$ and $\nu$, with slope $\frac{h}{e}$ and y-intercept $-\frac{\phi}{e}$. The figure shows parallel lines for $M_1$ and $M_2$, indicating the same slope, which is $\frac{h}{e}$. So, Statement I is correct.

Statement II: The statement claims that $M_2$ will emit photoelectrons of greater kinetic energy for the same incident frequency. This means $K_{max, M_2} > K_{max, M_1}$. Since $K_{max} = h\nu - \phi$, this implies $h\nu - \phi_{M_2} > h\nu - \phi_{M_1}$, which simplifies to $\phi_{M_2} < \phi_{M_1}$. From the graph, the threshold frequency for $M_1$ ($\nu_{0, M_1}$) is less than the threshold frequency for $M_2$ ($\nu_{0, M_2}$). Since $\phi = h\nu_0$, we have $\phi_{M_1} < \phi_{M_2}$. This contradicts the condition $\phi_{M_2} < \phi_{M_1}$ required for Statement II to be true. Therefore, Statement II is incorrect.