JEE Main 2014PhysicsAtomic PhysicsMediumMCQ

JEE Main 2014 — Atomic Physics Question with Solution

JEE Main 2014 (06 Apr)

Question

The radiation corresponding to 3 → 2  transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 1 0 - 4 T . If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

Choose an option

Show full solutionCorrect option: B
Correct answer
B1.1 eV

Step-by-step explanation

We know that

R = m υ Bq

∴    υ = Bqr m

∴    KE = 1 2 m v 2

= B 2 q 2 R 2 2m

= 0.79 eV

Now from photoelectric equation

E = ϕ + KE

∴    ϕ = E - KE

          = 1.89 - 0.79

          = 1.1 eV

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About this question

This is a previous-year question from JEE Main 2014, 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.