The first member of the Paschen series in the | vedclass

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(C) 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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(C) 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 $n = 4, 5, 6, \dots$.For the first member of the Paschen series,$n = 4$:$\frac{1}{\lambda_1} = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = \frac{7R}{144}$.Given $\lambda_1 = 18,800 \,\mathring{A}$,we have $R = \frac{144}{7 	imes 18,800} \,\mathring{A}^{-1}$.For the short wavelength limit (series limit),$n = \infty$:$\frac{1}{\lambda_{\infty}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = \frac{R}{9}$.Therefore,$\lambda_{\infty} = \frac{9}{R} = 9 	imes \left( \frac{7 	imes 18,800}{144} ight) = \frac{63 	imes 18,800}{144} = 8,225 \,\mathring{A}$.

The first member of the Paschen series in the hydrogen spectrum has a wavelength of $18,800 \,\mathring{A}$. The short wavelength limit of the Paschen series is ...... $\mathring{A}$.

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