The energy (in J) released when an excited el | vedclass

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Question

(A) The energy released during an electronic transition is given by the Rydberg formula: $E = 2.18 	imes 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n

Vedclass · Accepted Answer

$2.091 	imes 10^{-18}$

Vedclass · Answer

(A) The energy released during an electronic transition is given by the Rydberg formula: $E = 2.18 	imes 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ J$ Here,$n_1 = 1$ (ground state) and $n_2 = 5$ (excited state). Substituting the values: $E = 2.18 	imes 10^{-18} \left( \frac{1}{1^2} - \frac{1}{5^2} ight)$ $E = 2.18 	imes 10^{-18} \left( 1 - \frac{1}{25} ight)$ $E = 2.18 	imes 10^{-18} 	imes (1 - 0.04)$ $E = 2.18 	imes 10^{-18} 	imes 0.96$ $E = 2.0928 	imes 10^{-18} \ J$ The closest value is $2.091 	imes 10^{-18} \ J$.

The energy (in $J$) released when an excited electron of the $5^{th}$ orbit of a hydrogen atom returns to its ground state is

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