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

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$D < A < C < B$

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(D) The energy of a spectral line is given by $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$. For hydrogen $(Z=1)$, $\Delta E \propto (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.$A$. Paschen ($1^{st}$ line)$n_1=3, n_2=4 ightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.0486$$B$. Balmer ($2^{nd}$ line)$n_1=2, n_2=4 ightarrow \Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$$C$. Paschen ($3^{rd}$ line)$n_1=3, n_2=6 ightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{36}) = \frac{3}{36} = 0.0833$$D$. Bracket ($4^{th}$ line)$n_1=4, n_2=8 ightarrow \Delta E \propto (\frac{1}{16} - \frac{1}{64}) = \frac{3}{64} = 0.0468$Comparing the values: $0.0468 (D) Therefore, the correct ascending order is $D < A < C < B$.

Consider the following spectral lines for atomic hydrogen:
$A$. First line of Paschen series
$B$. Second line of Balmer series
$C$. Third line of Paschen series
$D$. Fourth line of Bracket series
The correct arrangement of the above lines in ascending order of energy is:

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Consider the following spectral lines for atomic hydrogen:$A$. First line of Paschen series$B$. Second line of Balmer series$C$. Third line of Paschen series$D$. Fourth line of Bracket seriesThe correct arrangement of the above lines in ascending order of energy is: