The extreme wavelengths of the Paschen series | vedclass

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(B) 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

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$0.818\,\mu m$ and $1.89\,\mu m$

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(B) 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 $n_2 = 4, 5, 6, \dots, \infty$.For the maximum wavelength (minimum energy transition),we take $n_1 = 3$ and $n_2 = 4$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16 - 9}{144} ight) = \frac{7R}{144}$.$\lambda_{\max} = \frac{144}{7R} = \frac{144}{7 	imes 1.097 	imes 10^7 \, 	ext{m}^{-1}} \approx 1.875 	imes 10^{-6} \, 	ext{m} \approx 1.89 \,\mu m$.For the minimum wavelength (maximum energy transition),we take $n_1 = 3$ and $n_2 = \infty$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = \frac{R}{9}$.$\lambda_{\min} = \frac{9}{R} = \frac{9}{1.097 	imes 10^7 \, 	ext{m}^{-1}} \approx 0.820 	imes 10^{-6} \, 	ext{m} \approx 0.818 \,\mu m$.Thus,the extreme wavelengths are $0.818 \,\mu m$ and $1.89 \,\mu m$.

The extreme wavelengths of the Paschen series are

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