# The angular momentum of an electron in 2nd orbit of an atom according to Bohr theory is

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The angular momentum of an electron in 2nd 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 = $\frac{h}{4\pi}$

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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 kth orbit $\frac{kh}{2\pi}$

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