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