Question

An electron in hydrogen atom stays in its second orbit for \(10^{-8}\) s. How many revolutions will it make around the nucleus in that time? [ Given : \(e = 1.6 \times 10^{-19}\) C, \(m = 9.1 \times 10^{-31}\) kg ]

Hint:

To find revolutions, you need the frequency (revolutions per second). Calculate the time for one revolution (Period, \(T\)) by finding the orbit's circumference (\(2\pi r_n\)) and dividing by the electron's speed (\(v_n\)). Then, divide the total time by the period.

Answer 1

First, we need to find the velocity (\(v_n\)) and radius (\(r_n\)) of the electron in the second orbit (\(n=2\)) of a hydrogen atom.
The velocity of an electron in the \(n^{th}\) orbit is \(v_n = \frac{e^2}{2\epsilon_0 h n}\), and the radius is \(r_n = \frac{\epsilon_0 h^2 n^2}{\pi m e^2}\).
A simpler approach is to calculate the time period of one revolution. The time period \(T_n\) of revolution in the \(n^{th}\) orbit is given by \(T_n = \frac{2\pi r_n}{v_n}\).
It can be shown that the angular frequency is \(\omega_n = \frac{\pi m e^4}{2\epsilon_0^2 h^3 n^3}\).
The frequency of revolution is \(f_n = \frac{\omega_n}{2\pi} = \frac{m e^4}{4\epsilon_0^2 h^3 n^3}\).
For the second orbit, \(n=2\).
\(f_2 = \frac{(9.1 \times 10^{-31}) (1.6 \times 10^{-19})^4}{4 (8.85 \times 10^{-12})^2 (6.63 \times 10^{-34})^3 (2)^3}\)
This is complex. Let's use a simpler known relation. The speed of the electron in the first orbit is \(v_1 \approx c/137 \approx 2.19 \times 10^6 \, m/s\).
The speed in the \(n^{th}\) orbit is \(v_n = v_1/n\). So, \(v_2 = v_1/2 \approx 1.095 \times 10^6 \, m/s\).
The radius of the first orbit is the Bohr radius, \(r_1 \approx 0.529 \times 10^{-10} \, m\).
The radius of the \(n^{th}\) orbit is \(r_n = r_1 n^2\). So, \(r_2 = r_1 (2)^2 = 4r_1 \approx 2.116 \times 10^{-10} \, m\).
The time period of one revolution in the second orbit is: \[ T_2 = \frac{2\pi r_2}{v_2} = \frac{2 \pi (2.116 \times 10^{-10})}{1.095 \times 10^6} \approx 1.21 \times 10^{-15} \, s \] The total time the electron stays in the orbit is \(t_{total} = 10^{-8} \, s\).
The number of revolutions is the total time divided by the time for one revolution: \[ \text{Number of revolutions} = \frac{t_{total}}{T_2} = \frac{10^{-8}}{1.21 \times 10^{-15}} \approx 8.26 \times 10^6 \] The electron will make approximately \(8.3 \times 10^6\) revolutions.