JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 12)
Given below are two statements :
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Statement I : Figure shows the variation of stopping potential with frequency $$(v)$$ 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 : $$\mathrm{M}_2$$ 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.
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 correct
Both Statement I and Statement II are incorrect
Explanation
$$\begin{aligned}
& V_0=\frac{h}{e} f-\frac{h}{e} f_0 \\
& \Rightarrow \text { Slope }=\frac{h}{e} \text { (S-I is correct) } \\
& \because\left(f_0\right)_1<\left(f_0\right)_2 \\
& \therefore \quad(\mathrm{S}-I I \text { is incorrect) } \\
& \therefore \quad \text { Option (3) is correct. }
\end{aligned}$$
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