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 $f/f_4$ is found to be $18/5m$. State the value of $m$.

Answer 1

Energy difference between $n=4$ and $n=5$ orbit of hydrogen atom, $\Delta E=13.6(\frac{1}{4^2}-\frac{1}{5^2})eV=0.306eV=0.49\times {10}^{-19}J$

Frequency of the photon emitted due to that transition is given by: $\Delta E=hf$

$\therefore 0.49\times {10}^{-19}=6.626\times {10}^{-34}f⟹f=7.395\times {10}^{13}{s}^{-1}$

Now frequency of the electron orbiting in $nth$ orbit, $f_n=\frac{6.56\times {10}^{15}}{n^3}$

$\therefore$ $f_4=\frac{6.56\times {10}^{15}}{4^3}=1.025\times {10}^{14}{s}^{-1}$

Thus, $\frac{f}{f_4}=\frac{7.389\times {10}^{13}}{1.025\times {10}^{14}}=0.721$

$\therefore \frac{18}{5m}=0.721⟹m=5$