JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2026
- Easy
- MCQ
The energy of an electron in an orbit of the Bohr's atom is \(-0.04 E _0 eV\) where \(E _0\) 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 _________ :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
Angular momentum \(L =\frac{ nh }{2 \pi}\)
\(n =\frac{2 \pi L}{ h }\)
Energy \(E =-\frac{13.6}{ n ^2} \cdot z ^2\)
\(E \Rightarrow-\frac{ E _0}{ n ^2}=-0.04 E _0\)
\(n ^2=25, n =5\)
\(n =\frac{2 \pi L}{ h }\)
Energy \(E =-\frac{13.6}{ n ^2} \cdot z ^2\)
\(E \Rightarrow-\frac{ E _0}{ n ^2}=-0.04 E _0\)
\(n ^2=25, n =5\)
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