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Find the ratio between the wavelength of the most energetic' spectral lines in the Balmer and Paschen series of the hydrogen spectrum. |
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Answer» \(\frac1{\lambda}=R(\frac1{n_1^2}-\frac1{n_2^2})\) R → Rydburg constant Balmer series n1 = 2 n2 = \(\infty\) for highest energy n2 → \(\infty\) \(\frac1{\lambda_B}=R(\frac1{2^2}-\frac1{\infty})\) \(\frac1{\lambda_B}=R(\frac1{4})\) \(\frac1{\lambda_B}=R(\frac R{4})\) \({\lambda_B}=\frac4{R}\) for paschen series n1 = 3, n2 = \(\infty\) \(\frac1{\lambda_p}=R(\frac{1}{(3^2)}-\frac1{(\infty^2)})\) \(\frac1{\lambda_p}=\frac{R}{p}\) \(\lambda_p=\frac{q}R,\) \(\frac{\lambda_b}{\lambda_p}=\cfrac{\frac4R}{\frac{q}R}\) ⇒ \(\frac4R\times\frac{R}q\) ⇒ 4 : q |
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