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Physics · 2025 · 2 marks
CBSE 2025 · Region 4 · Set 2 · Q20
Find the ratio of minimum to maximum wavelength of radiations emitted when electron jumps from higher energy state into ground state of hydrogen atom.
Find the ratio of minimum to maximum wavelength of radiations emitted when electron jumps from higher energy state into ground state of hydrogen atom.
Marking-scheme solution
$\displaystyle \frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$
$\displaystyle \frac{1}{\lambda_{\max }}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$
$\displaystyle \lambda_{\max }=\frac{4}{3 R}$
$\displaystyle \frac{1}{\lambda_{\min }}=R\left[\frac{1}{1^{2}}-\frac{1}{\infty}\right]$
$\displaystyle \lambda_{\min }=\frac{1}{R}$
$\displaystyle \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{3}{4}$
For $\displaystyle \lambda_{\min }$: $\displaystyle n_{1}=1, n_{2}=\infty$
$\displaystyle E_{2}-E_{1}=\frac{h c}{\lambda_{\min }}$
$\displaystyle 0-(-13.6)=\frac{h c}{\lambda_{\min }}$
$\displaystyle \lambda_{\min }=\frac{h c}{13.6}$
For $\displaystyle \lambda_{\max }$: $\displaystyle n_{1}=1, n_{2}=2$
$\displaystyle \lambda_{\max }=\frac{h c}{-3.4-(-13.6)}$
$\displaystyle \lambda_{\max }=\frac{h c}{10.2}$
$\displaystyle \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{3}{4}$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.