The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second…
The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to $2^{\text {nd }}$ level is respectively
$4: 1$
$2: 1$
$5: 1$
$3: 1$
Solution
The energy of photons is given by $E=\operatorname{Rhc}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
where, $\mathrm{R}$ is Rydberg constant, $\mathrm{h}$ is Planck's constant and $\mathrm{c}$ is the speed of light.
(i) Energy of photon produced from second to first energy level,
$\mathrm{E}_{1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\mathrm{E}_{1}=\frac{3}{4} \mathrm{Rhc}$
(ii) Energy of photon produced from highest energy level (i.e., \infty) to second level,
$\begin{array}{l}
\mathrm{E}_{2}=\operatorname{Rhc}\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{1}{4} \mathrm{Rhc} \\
\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{4} \mathrm{Rhc}}{\frac{1}{4} \mathrm{Rhc}}=\frac{3}{1} \text { or } 3: 1
\end{array}$