Question

Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is $2.14 \times {10^{ - 10}}m$ and $1.09 \times {10^6}m/s$ respectively. $\left[ {\pi = 3.14} \right]$

Answer 1

The Bohr model assumed that the electrons orbits the nucleus is in the fixed energy. In orbits the nucleus present in the higher energy levels. The electron returns in the lower energy then, the electron releases their energy in the form of the light.

Complete step by step solution:
In simple terms, frequency tells us the number of the waves that passed a fixed place in a given time.
The frequency of the revolution is the revolution per minute is the number of turns in one minute.
The Bohr orbit is defined as the circular orbit around the nucleus of the atom.
Now, we know the distance covered by the orbit is the circumference of the circle that is $2\pi r$
$\begin{aligned} & Speed = \dfrac{{\text{distance}}}{{\text{time}}} \\\ & \Rightarrow Speed= \dfrac{{2\pi r}}{T} \\\ \end{aligned}$
There is a rotational motion so
$v = \omega r$
We know, the relation between angular velocity and the frequency
So, $v = (2\pi f)r$
With the help of this expression we find the value of frequency
$f = \dfrac{v}{{2\pi r}}$--- (A)
The frequency of revolution of the electron in the second Bohr orbit
We know the radius and speed of the electron.
Given, radius, r = $2.14 \times {10^{ - 10}}m$,
Velocity, v = $1.09 \times {10^6}m/s$
Put the value in equation (A)
$\begin{aligned} & f = \dfrac{v}{{2\pi r}} \\\ & \Rightarrow f = \dfrac{{\left( {1.09 \times {{10}^6}} \right)}}{{\left( {2 \times 3.14 \times 2.14 \times {{10}^{ - 10}}} \right)}} \\\ & \Rightarrow f = \dfrac{{\left( {1.09 \times {{10}^6}} \right)}}{{\left( {13.4392 \times {{10}^{ - 10}}} \right)}} \\\ & \Rightarrow f = 0.081106 \times {10^{16}}Hz \\\ \end{aligned}$

Here’s the frequency of revolution of an electron in Bohr’s 2nd orbit.

Note: Electrons are quantum objects with properties like waves; Electrons always vibrate at some frequency. In a stable atomic state, the electron does not move in the sense of waving in space. In the time of vibration the orbital electron does move.