Question

What is the frequency of revolution of electron present in $2^{nd}$ Bohr's orbit of H-atom?

• A. $5.28\times {10}^{5}KHz$
• B. $8.5\times {10}^{16}Hz$
• C. $7.82\times {10}^{15}Hz$
• D. $8.13\times {10}^{16}Hz$

Answer 1

The correct option is D $8.13\times {10}^{16}Hz$

Frequency of the revolution is $6.55\times$ ${10}^1$${}^5$ $\frac{z^2}{n^3}$

Z = atomic number

n = number of orbit

So. Z= 1 and N = 2

Frequency of revolution is $6.55\times$ ${10}^1$${}^5$ $\times \frac{1}{8}$

= $8.13\times$ ${10}^1$${}^6$ Hz