Determine the frequency of revolution of an electron in the second Bohr orbit in hydrogen atom.

Explanation

The frequency of revolution, `f propto 1/n^(3), therefore (f_(2)/f_(1)) = 1^(3)/2^(3)`
`therefore f_(2) = f_(1)/8 = (6.576 xx 10^(15))/(8) = 8.22 xx 10^(14) Hz`