Question

The angular momentum of an electron in 2^nd orbit of an atom according to Bohr theory is 
1. Angular momentum = \(\frac{h}{\pi}\)
2. Angular momentum = \(\frac{2h}{\pi}\)
3. Angular momentum = \(\frac{3h}{2\pi}\)
4. Angular momentum = \(\frac{h}{4\pi}\)

Answer 1

Correct Answer - Option 1 : Angular momentum = \(\frac{h}{\pi}\)

The correct answer is option 1) i.e. Angular momentum = \(\frac{h}{\pi}\)

CONCEPT:

• Angular momentum of an electron in kth orbit:
  • Bohr’s atomic model suggests that the angular momentum of an electron orbiting around the nucleus of an atom is quantized.
  • Electrons transit only in those orbits where the angular momentum of an electron is an integral multiple of h/2, where h is the Planck's constant.
  • According to de Broglie, a moving electron in its circular orbit behaves like a particle-wave.
  • Thus standing waves are produced when the total distance travelled 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.

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: 

\(2πr_k = \frac{kh}{mv_k}\) ⇒\(mv_kr_k =\frac{kh}{2\pi}\) ⇒\(m\omega_k =\frac{kh}{2\pi}\)

∴ Angular momentum = \(m\omega_k =\frac{kh}{2\pi}\)

EXPLANATION:

Angular momentum of k^th orbit = \(\frac{kh}{2\pi}\)

In the 2^nd orbit, angular momentum = \(​​\frac{2h}{2\pi} =\frac{h}{\pi}\)