Question

A Bohr hydrogen atom undergoes a transition $n=5 \rightarrow n=4$ 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)

$E_{n}=-\frac{m z^{2} e^{4}}{8 \epsilon _{0}^{2} n^{2} h^{2}}$
So, $hv=+\frac{m z^{2} e^{4}}{8 \epsilon _{0}^{2} h^{2}}\left[\frac{1}{16} - \frac{1}{25}\right]$
$\therefore $ $v=\frac{m z^{2} e^{4}}{8 \epsilon _{0}^{2} h^{3}}\left[\frac{9}{16 \times 25}\right]$ $...\left(\right.1\left.\right)$
And frequency $v_{4}=\frac{1}{T}=\frac{v}{2 \pi r}$
$=\left(\frac{Z e^2}{2 \varepsilon_0 n h}\right) \frac{1}{2 \pi}\left(\frac{\pi m z e^2}{\varepsilon_0 h^2 n^2}\right)=\frac{z^2 e^4 m}{4 \varepsilon_0^2 n^3 h^3} \ldots....(2)$
$\therefore \frac{v}{v_{4}}=\frac{18}{25}=0.72$