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Question Number 77230 by Umar last updated on 04/Jan/20

$$\mathrm{An}\mathrm{astromer}\mathrm{finds}\mathrm{a}\mathrm{new}\mathrm{absorption}$$$$\mathrm{line}\mathrm{with}\lambda =164.1\mathrm{nm}\mathrm{in}\mathrm{the}\mathrm{ultraviolet}$$$$\mathrm{region}\mathrm{of}\mathrm{the}{\mathrm{sun}}^{\prime }\mathrm{s}\mathrm{continuous}\mathrm{spectrum}.$$$$\mathrm{He}\mathrm{attributes}\mathrm{the}\mathrm{line}\mathrm{to}{\mathrm{hydrogen}}^{\prime }\mathrm{s}$$$$\mathrm{Layman}\mathrm{series}.\mathrm{Is}\mathrm{he}\mathrm{right}?\mathrm{Justify}$$$$\mathrm{your}\mathrm{answer}.$$$$$$$$\mathbf{please}\mathbf{help}.$$

Answered by MrGaster last updated on 22/Jan/25

$$\frac{1}{164.1\times {10}^{-9}}-1.097\times {10}^7(1-\frac{1}{n_2^2})$$$$n_2^2=\frac{1}{1-\frac{1}{1.097\times 164.1\times {10}^{-2}}}$$$$n_2^2=\frac{1}{1-\frac{1}{180.5577}}$$$$n_2^2=\frac{1}{\frac{180.5577-1}{180.5577}}$$$$n_2^2=\frac{180.5577}{179.5577}$$$$n_2^2=1.00598$$$$\lambda =164.1\left(\mathrm{false}\right)$$

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