CBSE 2023 · Region 1 · Set 2 · Q25 · 2 marks
Calculate the wavelength of the second line of Lyman series in a spectrum of hydrogen atom. (Take Rydberg constant, $\displaystyle \mathrm{R}=1 \cdot 1 \times 10^{7} \mathrm{~m}^{-1}$ )
Marking-scheme solution
\(\displaystyle \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}{ }^{2}}-\frac{1}{\mathrm{n}_{2}{ }^{2}}\right]\) For second line of Lyman series \(\displaystyle \mathrm{n}_{1}=1, \mathrm{n}_{2}=3\) \(\displaystyle \frac{1}{\lambda}=1.1 \times 10^{7}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]\) \(\displaystyle =1.1 \times 10^{7}\left[1-\frac{1}{9}\right]\) \(\displaystyle =1.1 \times 10^{7} \times \frac{8}{9}\)
AtomsThe Line Spectra of the Hydrogen AtomApplyvery_short_answereasy
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.