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JEE MAIN 2026
21-1-2026 S2
Question
The energy of an electron in an orbit of the Bohr's atom is $-0.04 E_0 \mathrm{eV}$ where $\mathrm{E}_0$ 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 $\frac{2 \pi L}{h}$ is $\_\_\_\_$ :
Select the correct option:
A
2
B
4
C
5
D
6
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Easy
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