Q4 / 75 · JEE Main 2026 (28 January Shift 1) · Chemistry
The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral lines belonging to Balmer series.
($\mathrm{R}=$ Rydberg constant)
($\mathrm{R}=$ Rydberg constant)
- A$\frac{7 \mathrm{R}}{144}, \frac{3 \mathrm{R}}{16}, \frac{16 \mathrm{R}}{255}$
- B$\frac{5 \mathrm{R}}{36}, \frac{8 \mathrm{R}}{9}, \frac{15 \mathrm{R}}{16}$
- C$\frac{3 \mathrm{R}}{4}, \frac{3 \mathrm{R}}{16}, \frac{7 \mathrm{R}}{144}$
- Correct answerD$\frac{5 \mathrm{R}}{36}, \frac{3 \mathrm{R}}{16}, \frac{21 \mathrm{R}}{100}$
Solution
For the Balmer series, transitions end at n=2.
Using the Rydberg formula $\bar{\nu} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R\left(\frac{1}{4} - \frac{1}{n^2}\right)$, we calculate wavenumbers for transitions from different upper levels:
From n=3: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
From n=4: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$
From n=5: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{25}\right) = \frac{21R}{100}$
Option (4) contains exactly these three wavenumbers corresponding to the Balmer series transitions.
Using the Rydberg formula $\bar{\nu} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R\left(\frac{1}{4} - \frac{1}{n^2}\right)$, we calculate wavenumbers for transitions from different upper levels:
From n=3: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
From n=4: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$
From n=5: $\bar{\nu} = R\left(\frac{1}{4} - \frac{1}{25}\right) = \frac{21R}{100}$
Option (4) contains exactly these three wavenumbers corresponding to the Balmer series transitions.
Hence, the correct answer is option: 'D'.