According to Bohr's theory, the angular momentum of an electron in $$5^{th}$$ orbit is

Bohr postulated that the angular momentum $$L$$ of an electron in a stationary orbit is quantised according to

$$L = n \dfrac{h}{2\pi}$$

where

• $$n = 1,2,3,\dots$$ is the principal quantum number (orbit number).
• $$h$$ is Planck’s constant.

For the $$5^{\text{th}}$$ orbit we have $$n = 5$$, therefore

$$L = 5 \dfrac{h}{2\pi} = \dfrac{5}{2}\dfrac{h}{\pi} = 2.5\dfrac{h}{\pi}$$

Hence the angular momentum of the electron in the fifth Bohr orbit is $$2.5\dfrac{h}{\pi}$$.

Option D which is: $$2.5\dfrac{h}{\pi}$$