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JEE Main 20241 FebruaryMorning Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2024 · 1 February · Morning Shift · Physics · Atomic Physics

The minimum energy required by a hydrogen atom in ground state to...

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Options

  1. A1.5eV1 . 5eV
  2. B13.6eV13 . 6eV
  3. C1.9eV1 . 9eV
  4. D12.1eV12 . 1eV

Correct answer

D. 12.1eV12 . 1eV

Step-by-step solution

For Balmer series, transition is from n=3,4,5,6,etcn = 3, 4, 5, 6, etc to n=2n = 2.

For transition to n=2n = 2, the minimum state to transition from is n=3n = 3. 

Therefore, energy required \DeltaE=E3−E1=−13.69−(−13.6)=12.1eV\DeltaE = {E}_{3} - {E}_{1} = - \dfrac{13 . 6}{9} - \left(- 13 . 6\right) = 12 .1 eV

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