Question:
Given below are two statements:
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.
Given below are two statements:
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.
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
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The Correct Option is A
Solution and Explanation
The stopping potential (\(V_0\)) is related to frequency (\(\nu\)) by the equation:
\[ eV_0 = h\nu - \phi \implies V_0 = \frac{h}{e}\nu - \frac{\phi}{e} \]
The slope of the graph gives \(\frac{h}{e}\), confirming Statement-I. However, \(M_2\) has a higher work function, meaning that for the same incident frequency, the kinetic energy of emitted photoelectrons will be lower. Therefore, Statement-II is incorrect.
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