

InterviewSolution
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Angular momentum of an electron is an integral multiple of:1. \(\frac h {mc}\)2. \(\frac h {3\pi}\)3. \(\frac h {4\pi}\)4. \(\frac h {2\pi}\) |
Answer» Correct Answer - Option 4 : \(\frac h {2\pi}\) Concept: Bohr Model of Hydrogen Atom
This gives the relation:\(2πr_k = \frac{kh}{mv_k}\)= kλ ----(1) Where rkis the radius of the kthorbit, andλ is the wavelength. Also Thede Broglie wavelengthis given by: λ =\(\frac{h}{mv_k}\) ----(2) Where vkis the velocity of the electron in kthorbit. From (1) and (2) we get: \(2πr_k = \frac{kh}{mv_k}\) ⇒\(mv_kr_k =\frac{kh}{2π}\)⇒\(m\omega_k =\frac{kh}{2π}\) ∴Angular momentum=\(m\omega_k =\frac{kh}{2π}\) Explanation: So, the angular momentum is given by \(L =m\omega_k = mvr =\frac{kh}{2π}\) k and h are constant So, we can say that L∝ h / 2π So,\(\frac h {2\pi}\)is the correct option. |
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