Question

A Bohr 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 $\frac{25f}{9f_4}$ is

Answer 1

$E_n=-\frac{m{Z}^2{e}^4}{8{{\epsilon }_0}^2{n}^2{h}^2}\Rightarrow hf=\frac{m{Z}^2{e}^4}{8{{\epsilon }_0}^2{h}^2}\left(\frac{1}{16}-\frac{1}{25}\right]\therefore f=\frac{9m{Z}^2{e}^4}{3200{{\epsilon }_0}^2{h}^3}$
And, frequency $f_4=\frac{e^{4}m{z}^2}{4{{\epsilon }_0}^2{h}^3{n}^3}=\frac{e^{4}m{z}^2}{\left(4^3\right)\times 4{{\epsilon }_0}^2{h}^3}=\frac{e^{4}m{z}^2}{256{{\epsilon }_0}^2{h}^3}$$\Rightarrow \frac{f}{f_4}=\frac{9}{3200}\times \frac{256}{1}=\frac{18}{25}$
So, $\frac{25f}{9f_4}=2$