Question

According to Bohr theory, the angular momentum of electron in ${5^{{\text{th}}}}$ orbit is
A.$25\dfrac{h}{\pi }$
B.$1\dfrac{h}{\pi }$
C.$10\dfrac{h}{\pi }$
D.$2.5\dfrac{h}{\pi }$

Answer 1

Hint:To answer this question, you must recall the postulates given by Bohr in Bohr’s atomic theory. In the third postulate, it is stated that the angular momentum of an electron in any orbit is quantized.
Formula used: $n\dfrac{h}{{2\pi }}$
Where, $n$ denotes any integer
And, $h$ denotes the Planck’s constant.

Complete step by step answer:
The Bohr’s model of atom was among the first atomic models that successfully explained the position of the absorption and emission lines shown for hydrogen atom and hydrogen like species. He formulated three postulates for the explanation of his model.
The electrons in an atom revolve around the nucleus in fixed circular energy levels called orbitals. Each orbital has a fixed energy.
These energy levels are denoted by integral values termed as quantum numbers. This quantum number starts from the nucleus and outwards with n=1 as the lowest energy level. An electron in its lowest energy level is said to be present in its ground state. The electrons move from a lower energy level to higher energy level by absorbing energy and an electron from the higher energy level moves to a lower energy level by losing energy.
The third postulate of Bohr's theory states that the angular momentum of an electron revolving around the nucleus in a fixed energy level is quantized. It is said to be an integral multiple of $\dfrac{h}{{2\pi }}$with the integer being equal to the energy level of the electron.
So the angular momentum of electron in the 5th orbit is given as
$mvr = n\dfrac{h}{{2\pi }} = 5\dfrac{h}{{2\pi }}$
$\therefore mvr = 2.5\dfrac{h}{\pi }$

Hence, the correct answer is D.

Note:
Although Bohr’s model of atom was successful in explaining a lot of concepts, its applications were limited only to hydrogen and hydrogen- like species containing one electron.