Calculate The Shortest And Longest Wavelengths In Hydrogen Spectrum Of Lyman Series.

For Lyman series, n_{1} = 1

For shortest wavelength in Lyman series \(n_{1} = \infty\)

So, \(\frac{1}{\lambda} = R_{H} [\frac{1}{1^{2}} – \frac{1}{\infty^{2}}] = R_{H}\) \(\lambda = \frac{1}{109678} = 9.117 * 10^{-6}cm\) \(\lambda = 911.7 {A^{\circ}}\) \(For longest wavelength in Lyman series, n_{2} = 2 \)

So, \(\frac{1}{\lambda} = R_{H} [\frac{1}{1^{2}} – \frac{1}{2^{2}}] = \frac{3}{4} R_{H} \)

or

\(\lambda = \frac{4}{3} * \frac{1}{R_{H}} \)

=\(\frac{4}{3 * 109678}\)

=\(1215.7 * 10^{-8}cm\)

=\(1215.7 {A^{\circ}}\)

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