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Emission and Absorption Spectra

If n 1 and n ...

Question

If $n_{1}$ and $n_{2}$ are the boundary value principal quantum numbers of a portion of spectrum of emission spectrum of H atom, determine the wavelength (in metre) corresponding to last line (longest $λ$ ). Given : $n_{1} + n_{2} = 7, n_{2} - n_{1} = 3, a n d R_{H} = 1.097 \times 10^{7} m^{- 1} .$ (Give your answer in multiple of $10^{- 6}$ )

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Solution

Here $n_{2} = 5 a n d n_{1} = 2$
So longest wavelength means least energy difference transition i.e. $n_{2} = 5 t o n_{1} = 4$
$\frac{1}{λ} = R_{H} (1)^{2} (\frac{1}{(4)^{2}} - \frac{1}{(5)^{2}})$
So $λ = 4 \times 10^{- 6}$ m

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