MP_JEE Main 2022_28-06-22_{S2} - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2022

28-06-22

Question

Let $K_1$ and $K_2$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_1$ and $\lambda_2$, respectively are incident on a metallic surface. If $\lambda_1=3 \lambda_2$ then:

Select the correct option:

$\mathrm{K}_1>\frac{\mathrm{K}_2}{3}$

) $K_1<\frac{K_2}{3}$

$\mathrm{K}_1=\frac{K_2}{3}$

$\mathrm{K}_2=\frac{\mathrm{K}_1}{3}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}-\phi=\mathrm{K}_1 \\ & \frac{\mathrm{hc}}{\lambda_2}-\phi=\mathrm{K}_2 \\ & \lambda_1=3 \lambda_2 \\ & 3 \mathrm{~K}_1=\frac{3 \mathrm{hc}}{\lambda_1}-3 \phi \\ & 3 \mathrm{~K}_1=\frac{\mathrm{hc}}{\lambda_2}-3 \phi \\ & 3 \mathrm{~K}_1=\mathrm{K}_2-2 \phi \\ & 3 \mathrm{~K}_1<\mathrm{K}_2 \\ & \mathrm{~K}_1<\frac{\mathrm{K}_2}{3}\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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