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JEE MAIN 2024
05-04-2024 S1
Question
Given below are two statements: 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 $\frac{\mathrm{h}}{\mathrm{e}}$, 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.
Select the correct option:
A
Statement-I is correct and Statement-II is incorrect.
B
Statement-I is incorrect but Statement-II is correct.
C
Both Statement-I and Statement-II are incorrect.
D
Both Statement-I and Statement-II are correct.
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & \mathrm{eV}_0=\mathrm{hv}-\phi \\ & \mathrm{V}_0=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{v}-\frac{\phi}{\mathrm{e}} \end{aligned} $$ $\mathrm{M}_2$ material has higher work function, so statement-(II) is incorrect.
Question Tags
JEE Main
Physics
Medium
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