Match the following for hydrogen atom (Z=1) : | vedclass

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(C) For a hydrogen atom,the energy difference between levels is $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.Since $\D

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$P-ii, Q-iii, R-i, S-iv$

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(C) For a hydrogen atom,the energy difference between levels is $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.Since $\Delta E = \frac{hc}{\lambda} = hv$,$\lambda$ is inversely proportional to $\Delta E$,and $v$ is directly proportional to $\Delta E$.$(P). \lambda_{\max }$ in Lyman series: Requires minimum $\Delta E$,which is $2 ightarrow 1$.$(Q). v_{\min }$ in Balmer series: Requires minimum $\Delta E$,which is $3 ightarrow 2$.$(R). \lambda_{\min }$ in Lyman series: Requires maximum $\Delta E$,which is $\infty ightarrow 1$.$(S). v_{\max }$ in Balmer series: Requires maximum $\Delta E$,which is $\infty ightarrow 2$.Therefore,the correct matching is $P-ii, Q-iii, R-i, S-iv$.

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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$(P). \lambda_{\max }$ 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	$(iv). \infty \rightarrow 2$

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$(iv). \infty \rightarrow 2$