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Physics · Atomic Physics

JEE Main 2026 — 21 January, Evening Shift — Question 40

The energy of an electron in an orbit of the Bohr's atom is −0.04E0eV-0.04 \mathrm{E}_{0} \mathrm{eV} where E0\mathrm{E}_{0} is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2π L h\frac{2 \pi \mathrm{~L}}{\mathrm{~h}} is ____\_\_\_\_ :

  1. Option A:

    2

  2. Option B:

    4

  3. Option C:

    5

    Correct
  4. Option D:

    6

Answer: C

Step-by-step solution

Angular momentum L=nh2π\mathrm{L}=\frac{\mathrm{nh}}{2 \pi} n=2π L h\mathrm{n}=\frac{2 \pi \mathrm{~L}}{\mathrm{~h}} Energy E=−13.6n2.z2\mathrm{E}=-\frac{13.6}{\mathrm{n}^{2}} . \mathrm{z}^{2}

E⇒−E0n2=−0.04E0n2=25,n=5\begin{aligned} & \mathrm{E} \Rightarrow-\frac{\mathrm{E}_{0}}{\mathrm{n}^{2}}=-0.04 \mathrm{E}_{0} & \mathrm{n}^{2}=25, \mathrm{n}=5 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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