Question

The angular speed of an electron in nth orbit of hydrogen.

Answer 1

Electrons are the negative subatomic particles of an atom. They possess circular motion in the orbits of electrons. The speed of an electron due to its rotation is known as angular speed. It is denoted by $ω_n$. Electrons when rotate do not radiate energy and do not fall into the nucleus of an atom.

Complete step-by-step answer:
In nth orbit the angular speed gives relation with linear speed as
$\mathop v\nolimits_{n = \mathop r\nolimits_n \mathop \omega \nolimits_n }$ where $r_n$ is the radius of atom and $v_n$ is the linear velocity of electron in nth orbital.

Now $\mathop \omega \nolimits_n = \dfrac{{\mathop v\nolimits_n }}{{\mathop r\nolimits_n }}$

But we know linear velocity of electrons is

$\mathop v\nolimits_n = \dfrac{{\mathop {ze}\nolimits^2 }}{{2\varepsilon \circ nn}}$

Where z= 1 for hydrogen, e is charged on an electron, E° is permeability, n is defined as the orbital number of hydrogen atom.
Now when we put the value of $\mathop v\nolimits_n$
$\mathop \omega \nolimits_n = \dfrac{{\mathop {ze}\nolimits^2 }}{{2\varepsilon \circ nn}} \propto \dfrac{1}{{\mathop r\nolimits_n }}$

This shows that
$\mathop \omega \nolimits_n \propto \dfrac{1}{n}$
means angular velocity of electron in nth orbital of hydrogen is inversely proportional to that of orbital number.

Note: Bohr told that while revolving electrons do not radiate energy and do not fall down into the nucleus. It revolves with constant speed. And its angular speed depends inversely on the orbital number. The closer to the nucleus the more it will possess angular speed and the more distance it has for the nucleus the less speed it will possess.