Chemistry Q52 – JEE-MAIN 2026

Home Chemistry Structure of AtomQuestion #600

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

Clue 1 of 3: Atomic Spectra Series

Recall the principal quantum numbers ( $n_{f}$ and $n_{i}$ ) for the first line (lowest energy transition) in the Balmer and Brackett series for the hydrogen atom.

🧠 Full Solution Path

Step 1: Identify Quantum Numbers for Each Series✦ Active

For the first line (lowest energy) of the Balmer series, the electron transitions from $n_{i} = 3$ to $n_{f} = 2$ . For the first line (lowest energy) of the Brackett series, the electron transitions from $n_{i} = 5$ to $n_{f} = 4$ .

Step 2: Apply the Rydberg Formula for Wavenumber○ Expand

The wavenumber $\bar{ν}$ for a hydrogen atom ( $Z = 1$ ) is given by the Rydberg formula: $\bar{ν} = R_{H} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ .

{\bar{ν}}_{B a l m e r} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = R_{H} (\frac{9 - 4}{36}) = \frac{5 R_{H}}{36}

{\bar{ν}}_{B r a c k e t t} = R_{H} (\frac{1}{4^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{1}{16} - \frac{1}{25}) = R_{H} (\frac{25 - 16}{400}) = \frac{9 R_{H}}{400}

Step 3: Calculate the Ratio of Wavenumbers○ Expand

Now, calculate the ratio of the wavenumber of the Balmer series' first line to that of the Brackett series' first line.

\frac{{\bar{ν}}_{B a l m e r}}{{\bar{ν}}_{B r a c k e t t}} = \frac{\frac{5 R_{H}}{36}}{\frac{9 R_{H}}{400}} = \frac{5}{36} \times \frac{400}{9} = \frac{5 \times 100}{9 \times 9} = \frac{500}{81}

The ratio is $500 : 81$ . To match the given options, we can express this as $5 : \frac{81}{100}$ , which simplifies to $5 : 0.81$ .

💡 Teacher's Secret Hint

Ensure to simplify the fraction correctly to match the format of the options.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Atomic Spectra Series

🧠 Full Solution Path