JEE Main 2019PhysicsAtomic PhysicsMediumMCQ

JEE Main 2019 — Atomic Physics Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

In a Frank - Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

Choose an option

Show full solutionCorrect option: A
Correct answer
A250 nm

Step-by-step explanation

When electron pass through the mercury vapor, it losses some of its energy. The loss in KE of electron =5⋅6-0.7eV =4.9 eV

∴ energy of radiation emitted = 4.9 eV

∴ wavelength of radiation, λ=1.24 × 1044.9A≈ 250 nm

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Atomic Physics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Atomic Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.