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

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Question

(A) For the Lyman series,the transition occurs to the $n_1 = 1$ orbit from $n_2 = 2, 3, 4, \dots$ levels.Using the Rydberg formula: $\frac{1}{\lambda}

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

Vedclass · Answer

(A) For the Lyman series,the transition occurs to the $n_1 = 1$ orbit from $n_2 = 2, 3, 4, \dots$ levels.Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $R \approx 1.097 	imes 10^7 	ext{ m}^{-1}$.For the longest wavelength (minimum energy transition),$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4}$.$\lambda_{\max} = \frac{4}{3R} \approx \frac{4}{3 	imes 1.097 	imes 10^7} \approx 1216 \ \mathring{A}$ (approx $1213 \ \mathring{A}$).For the shortest wavelength (maximum energy transition),$n_1 = 1$ and $n_2 = \infty$:$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R$.$\lambda_{\min} = \frac{1}{R} \approx \frac{1}{1.097 	imes 10^7} \approx 912 \ \mathring{A}$ (approx $910 \ \mathring{A}$).Thus,the shortest wavelength is $910 \ \mathring{A}$ and the longest is $1213 \ \mathring{A}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

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

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