Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1153 Upvotes
Verified Answer
The correct answer is:
$3: 1$
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,
$\begin{array}{l}
\mathrm{E}_{2}=\operatorname{Rhc}\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{1}{4} \mathrm{Rhc} \\
\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{4} \mathrm{Rhc}}{\frac{1}{4} \mathrm{Rhc}}=\frac{3}{1} \text { or } 3: 1
\end{array}$
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,
$\begin{array}{l}
\mathrm{E}_{2}=\operatorname{Rhc}\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{1}{4} \mathrm{Rhc} \\
\therefore \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{4} \mathrm{Rhc}}{\frac{1}{4} \mathrm{Rhc}}=\frac{3}{1} \text { or } 3: 1
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.