Question

A Bohr hydrogen atom undergoes a transition $n=5\to n=4$ and emits a photon of frequency $\upsilon$ . Frequency of circular motion of electron in $n=4$ orbit is $v_4$ . The ratio $\frac{\nu }{{\nu }_4}$ is

• A. $\frac{18}{25}$
• B. $\frac{16}{25}$
• C. $\frac{9}{25}$
• D. $\frac{8}{25}$

Answer 1

Answer: (A)

$E_n=-\frac{m{z}^2{e}^4}{8{ϵ}_0^2{n}^2{h}^2}$
So, $hv=+\frac{m{z}^2{e}^4}{8{ϵ}_0^2{h}^2}\left[\frac{1}{16}-\frac{1}{25}\right]$
$\therefore$ $v=\frac{m{z}^2{e}^4}{8{ϵ}_0^2{h}^3}\left[\frac{9}{16\times 25}\right]$ $...(1)$
And frequency $v_4=\frac{1}{T}=\frac{v}{2\pi r}$
$=\left(\frac{Z{e}^2}{2{\epsilon }_{0}nh}\right)\frac{1}{2\pi }\left(\frac{\pi mz{e}^2}{{\epsilon }_0{h}^2{n}^2}\right)=\frac{z^2{e}^{4}m}{4{\epsilon }_0^2{n}^3{h}^3}\dots ....(2)$
$\therefore \frac{v}{v_4}=\frac{18}{25}=0.72$