The ratio of the longest to the shortest wave | vedclass

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(A) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 3$ an

Vedclass · Accepted Answer

$\frac{144}{63}$

Vedclass · Answer

(A) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $n_1 = 3$ and $n_2 = 4, 5, 6, \dots$ The longest wavelength $(\lambda_{\max})$ occurs for the transition from $n_2 = 4$ to $n_1 = 3$: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = \frac{7R}{144} \implies \lambda_{\max} = \frac{144}{7R}$ The shortest wavelength $(\lambda_{\min})$ occurs for the transition from $n_2 = \infty$ to $n_1 = 3$: $\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = R \left( \frac{1}{9} - 0 ight) = \frac{R}{9} \implies \lambda_{\min} = \frac{9}{R}$ The ratio of the longest to the shortest wavelength is: $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{144/7R}{9/R} = \frac{144}{7 	imes 9} = \frac{144}{63}$

The ratio of the longest to the shortest wavelength emitted in the Paschen series of a hydrogen atom is:

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