# 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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