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(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} \right]$,where $n = 4, 5, 6, \d

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$820$

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(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} ight]$,where $n = 4, 5, 6, \dots$ To find the shortest wavelength,we consider the limit where $n = \infty$. Substituting $n = \infty$ into the formula: $\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{9} - 0 ight] = \frac{R}{9}$. Thus,$\lambda_{\min} = \frac{9}{R}$. Given the Rydberg constant $R \approx 1.097 	imes 10^7 \ m^{-1}$: $\lambda_{\min} = \frac{9}{1.097 	imes 10^7} \approx 8.204 	imes 10^{-7} \ m$. Converting to nanometers: $\lambda_{\min} \approx 820.4 \ nm \approx 820 \ nm$.

What is the shortest wavelength present in the Paschen series of spectral lines (in $nm$)?

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