Verified Solution • 2025–26 Session moderate

QJEEM21Ph4a50: JEE Main 2021 Physics - Class 12 JEEM21Ph4a50 Solution: A free electron of 2.6 eV energy collides with a H + ion. This…

A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 × 10−34 Js)

Multiple Choice Options

  • A 1.45 × 1016 MHz
  • B 0.19 × 1015 MHz
  • C 1.45 × 109 MHz
  • D 9.0 × 1027 MHz

Step-by-Step Verified Solution

For every large distance P.E. = 0

& total energy = 2.6 + 0 = 2.6 eV

Finally in first excited state of H atom total energy = −3.4 eV

Loss in total energy = 2.6 − (−3.4) = 6 eV

It is emitted as photon

λ=12406=206 nm

f=3×108206×10−9 = 1.45 × 1015 Hz

= 1.45 × 109 Hz

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring