The energy of an electron in the n^{th} orbit | vedclass

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(A) The Lyman series corresponds to transitions where the final state is $n_f = 1$. The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1

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$910 \, \mathring{A}, 1213 \, \mathring{A}$

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(A) The Lyman series corresponds to transitions where the final state is $n_f = 1$. The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$,where $R \approx 1.097 	imes 10^7 \, m^{-1}$.For the longest wavelength (minimum energy transition),we take $n_i = 2$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4}$.$\lambda_{\max} = \frac{4}{3R} \approx \frac{4}{3 	imes 1.097 	imes 10^7} \approx 1.215 	imes 10^{-7} \, m = 1215 \, \mathring{A}$ (approximately $1213 \, \mathring{A}$ depending on the value of $R$ used).For the shortest wavelength (maximum energy transition),we take $n_i = \infty$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R$.$\lambda_{\min} = \frac{1}{R} \approx \frac{1}{1.097 	imes 10^7} \approx 9.11 	imes 10^{-8} \, m = 911 \, \mathring{A}$ (approximately $910 \, \mathring{A}$).Thus,the shortest wavelength is $910 \, \mathring{A}$ and the longest is $1213 \, \mathring{A}$.

The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is expressed as $E_n = \frac{-13.6}{n^2} \, eV$. The shortest and longest wavelengths of the Lyman series will be:

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