Physics · Atomic Physics

JEE Main 2024 — 9 April, Shift 2 — Question 42

UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV . The maximum kinetic energy of ejected photoelectrons will be :

Option A:
4.13 eV
Option B:
1 eV
Correct
Option C:
3.13 eV
Option D:
7.26 eV

Answer: B

Step-by-step solution

$\mathrm{E}_{\text {photon }}=($ work function $)+$ K. $\mathrm{E}_{\max }$

\begin{aligned} & \therefore 4 \cdot 13=3 \cdot 13+\mathrm{K} \cdot \mathrm{E}_{\max } & \therefore \mathrm{K} \cdot \mathrm{E}_{\max }=1 \mathrm{eV} \end{aligned}

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect

A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to transition of electrons?

Answer: B

Step-by-step solution

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