Question:medium

Given below are two statements:
Fig
Statement-I: Figure shows the variation of stopping potential with frequency (v) for the two photosensitive materials M1 and M2. The slope gives value of \(\frac{h}{e}\) , where h is Planck's constant, e is the charge of electron.
Statement-II: M2 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.

Updated On: Jan 13, 2026
  • Statement-I is correct and Statement-II is incorrect.
  • Statement-I is incorrect but Statement-II is correct.
  • Both Statement-I and Statement-II are incorrect.
  • Both Statement-I and Statement-II are correct.
Show Solution

The Correct Option is A

Solution and Explanation

The stopping potential (\(V_0\)) is related to frequency (\(u\)) by the equation:

\[ eV_0 = hu - \phi \implies V_0 = \frac{h}{e}u - \frac{\phi}{e} \]

The slope of the graph, which represents \(\frac{h}{e}\), confirms Statement-I. Statement-II is incorrect because \(M_2\) has a higher work function, resulting in lower kinetic energy for emitted photoelectrons at the same incident frequency.

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