An electronic transition in a hydrogen atom r | vedclass

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(B) The energy of an electron in the $n^{	ext{th}}$ orbit is given by $E_n = -\frac{313.6 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom,$Z = 1$,so

Vedclass · Accepted Answer

$-313.6, -78.4$

Vedclass · Answer

(B) The energy of an electron in the $n^{	ext{th}}$ orbit is given by $E_n = -\frac{313.6 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom,$Z = 1$,so $E_n = -\frac{313.6}{n^2} \ kcal \ mol^{-1}$.The Lyman series corresponds to transitions ending at $n_1 = 1$. The $H_\alpha$ line in the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.For $n_1 = 1$,$E_1 = -\frac{313.6 	imes 1^2}{1^2} = -313.6 \ kcal \ mol^{-1}$.For $n_2 = 2$,$E_2 = -\frac{313.6 	imes 1^2}{2^2} = -\frac{313.6}{4} = -78.4 \ kcal \ mol^{-1}$.Thus,the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

An electronic transition in a hydrogen atom results in the formation of the $H_\alpha$ line of the hydrogen spectrum in the Lyman series. The energies associated with the electron in each of the orbits involved in the transition (in $kcal \ mol^{-1}$) are:

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