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(C) The energy change for a transition in a hydrogen atom is given by the formula: $\Delta E = 2.18 	imes 10^{-18} \ J 	imes (\frac{1}{n_1^2} - \fra

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$3.03 	imes 10^{-19} \ J \ atom^{-1}$

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(C) The energy change for a transition in a hydrogen atom is given by the formula: $\Delta E = 2.18 	imes 10^{-18} \ J 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Here,$n_1 = 2$ and $n_2 = 3$.$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{1}{2^2} - \frac{1}{3^2})$$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{1}{4} - \frac{1}{9})$$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{9-4}{36})$$\Delta E = 2.18 	imes 10^{-18} 	imes \frac{5}{36}$$\Delta E = 0.3027 	imes 10^{-18} \ J \ atom^{-1}$$\Delta E = 3.03 	imes 10^{-19} \ J \ atom^{-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	$(iv). \infty \rightarrow 2$

Calculate the energy released when an electron in a hydrogen atom transitions from the $3^{rd}$ energy level to the $2^{nd}$ energy level.

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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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