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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. \(\frac h {2\pi}\)

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Correct Answer - Option 4 : \(\frac h {2\pi}\)

Concept:

Bohr Model of Hydrogen Atom

  • Bohr proposed a model for the hydrogen atom in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen-like atom).
  • A moving electron in its circular orbit behaves like a particle-wave.
  • As a result, standing waves are produced and the total distance traveled by a wave is an integral number of wavelengths. 

​This gives the relation: \(2πr_k = \frac{kh}{mv_k}\) = kλ      ----(1)

Where rk is the radius of the kth orbit, and λ is the wavelength.

Also The de Broglie wavelength is given by: 

λ = \(\frac{h}{mv_k}\)      ----(2)

​Where vk is the velocity of the electron in kth orbit.

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 / 2 π 

So, \(\frac h {2\pi}\) is the correct option.

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