JEE Mains · Physics · STD 12 - 12. atoms
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 _________ :
- A 2
- B 4
- C 5
- D 6
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 \(E =-\frac{13.6}{ n ^2} \cdot z ^2\) \(E \Rightarrow-\frac{ E _0}{ n ^2}=-0.04 E _0\) \(n ^2=25, n =5\)
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