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Given below are two statements: Statement I: Figure shows the variation of stopping potential with frequency ( u) for the two photosensitive materials M₁ and M₂. The slope gives value of 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.

Asked in JEE Main 5th April 1st Shift 2024 · Work function from a graph

Figure: Work function from a graph
Answer: (2) Statement I is correct and Statement II is incorrect.

Step-by-step solution

Statement I: V₀=h/e u-φ/e, so the slope of either line is h/e. Correct.

Statement II: in the figure M₁ meets the u axis first, so u_0,M₁< u_0,M₂ and φ_M₁<φ_M₂.

At one frequency, Kₘₐₓ=h u-φ is larger for the smaller work function, which is M₁.

So Statement II is incorrect.

Why the other options are wrong

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