The smallest wavelength of the Lyman series i | vedclass

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(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Lyman series, the smallest wavele

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$1217$

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(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series, the smallest wavelength corresponds to $n_1 = 1$ and $n_2 = \infty$.$\frac{1}{\lambda_{L,min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R = \frac{1}{91} \ nm^{-1}$.For the Balmer series, the largest wavelength corresponds to $n_1 = 2$ and $n_2 = 3$.$\frac{1}{\lambda_{B,max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = \frac{1}{91} \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{1}{91} \left( \frac{5}{36} ight)$.$\lambda_{B,max} = \frac{91 	imes 36}{5} = 655.2 \ nm$.For the Paschen series, the largest wavelength corresponds to $n_1 = 3$ and $n_2 = 4$.$\frac{1}{\lambda_{P,max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = \frac{1}{91} \left( \frac{1}{9} - \frac{1}{16} ight) = \frac{1}{91} \left( \frac{7}{144} ight)$.$\lambda_{P,max} = \frac{91 	imes 144}{7} = 1872 \ nm$.The difference is $\Delta\lambda = \lambda_{P,max} - \lambda_{B,max} = 1872 - 655.2 = 1216.8 \ nm \approx 1217 \ nm$.

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

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