An electronic transition in a hydrogen atom r | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom,$Z = 1$.The Lyma

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$-313.6, -78.4$

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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom,$Z = 1$.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 $n_1 = 1$.Energy in the $n_1 = 1$ orbit: $E_1 = -\frac{313.52 	imes (1)^2}{(1)^2} = -313.52 \ kcal \ mol^{-1} \approx -313.6 \ kcal \ mol^{-1}$.Energy in the $n_2 = 2$ orbit: $E_2 = -\frac{313.52 	imes (1)^2}{(2)^2} = -\frac{313.52}{4} = -78.38 \ kcal \ mol^{-1} \approx -78.4 \ kcal \ mol^{-1}$.Thus,the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

$(P). \lambda_{\max }$ in Lyman series	$(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series	$(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series	$(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series	$(iv). \infty \rightarrow 2$

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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Match the following for hydrogen atom $(Z=1)$ :-
$(P). \lambda_{\max }$ in Lyman series $(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series $(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series $(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series $(iv). \infty \rightarrow 2$

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