The hydrogen spectrum consists of several spe | vedclass

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(B) The energy of a spectral line is given by $\Delta E = 13.6 	imes Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.For the Lyman series, the lo

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$54$

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(B) The energy of a spectral line is given by $\Delta E = 13.6 	imes Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.For the Lyman series, the lowest energy line $(L_1)$ corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\Delta E(L_1) = 13.6 	imes 1^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.For the Balmer series, the lowest energy line $(B_1)$ corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\Delta E(B_1) = 13.6 	imes 1^2 (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 	imes (\frac{1}{4} - \frac{1}{9}) = 13.6 	imes \frac{5}{36} 	ext{ eV}$.Given $\Delta E(L_1) = x 	imes \Delta E(B_1)$, we have $x = \frac{\Delta E(L_1)}{\Delta E(B_1)} = \frac{3/4}{5/36} = \frac{3}{4} 	imes \frac{36}{5} = \frac{27}{5} = 5.4$.To express $x$ as $x 	imes 10^{-1}$, we have $5.4 = 54 	imes 10^{-1}$.Therefore, the value is $54$.

List-$I$ (Spectral Series for Hydrogen)	List-$II$ (Spectral Region)
$A$. Lyman	$I$. Infrared region
$B$. Balmer	$II$. $UV$ region
$C$. Paschen	$III$. Infrared region
$D$. Pfund	$IV$. Visible region

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List-$I$ (Spectral Series for Hydrogen) List-$II$ (Spectral Region)

$A$. Lyman $I$. Infrared region

$B$. Balmer $II$. $UV$ region

$C$. Paschen $III$. Infrared region

$D$. Pfund $IV$. Visible region

Choose the correct answer from the options given below:

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