A Bohr's hydrogen atom undergoes a transition `n = 5 to n = 4` and emits a photon of frequency `f`. Frequency of circular motion of electron in `n = 4` orbit is `f_(4)`. The ratio `f//f_(4)` is found to be `18//5 m`. State the value of `m`.

Answer

5

Explanation

`E_(n) = - (m Z^(2) e^(4))/(8 epsilon_(0)^(2) n^(2) h^(2)), so hf = + (m Z^(2) e^(4))/(8 epsilon_(0)^(2)h^(2)) [(1)/(16) - (1)/(25)]`
`f = (m Z^(2) e^(4))/(8 epsilon_(0)^(2)h^(2)) [(9)/(16 xx 25)]` (i)
and frequency `f_(4) = (Z^(2) e^(4)m)/(8 epsilon_(0)^(2) n^(2) h^(2)) = (Z^(2) e^(4)m)/(8 epsilon_(0)^(2) (4)^(3) h^(3))` (ii)
`:. f//f_(4) = 18//25, so m = 5`