Match List I with List II. List I (Spect | vedclass

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$(A)$ The energy difference for a transition is given by $\Delta E = \frac{hc}{\lambda}$.Therefore, the wavelength is inversely proportional to the en

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$A-III, B-IV, C-II, D-I$

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$(A)$ The energy difference for a transition is given by $\Delta E = \frac{hc}{\lambda}$.Therefore, the wavelength is inversely proportional to the energy difference: $\lambda \propto \frac{1}{\Delta E}$.As the energy difference increases, the wavelength decreases.For the Balmer series $(n_1=2)$, the energy differences are:$(\Delta E)_{3-2} Consequently, the wavelengths follow the order:$\lambda_{3-2} > \lambda_{4-2} > \lambda_{5-2} > \lambda_{6-2}$.Matching the values:$A$ ($n_2=3$ to $n_1=2$) corresponds to $656.3 \, nm$ $(III)$.$B$ ($n_2=4$ to $n_1=2$) corresponds to $486.1 \, nm$ $(IV)$.$C$ ($n_2=5$ to $n_1=2$) corresponds to $434.1 \, nm$ $(II)$.$D$ ($n_2=6$ to $n_1=2$) corresponds to $410.2 \, nm$ $(I)$.Thus, the correct match is $A-III, B-IV, C-II, D-I$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

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