JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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)

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

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
$\frac{3 \mathrm{R}}{4}, \frac{3 \mathrm{R}}{16}, \frac{7 \mathrm{R}}{144}$
D
$\frac{5 \mathrm{R}}{36}, \frac{3 \mathrm{R}}{16}, \frac{21 \mathrm{R}}{100}$

Did you get this right?

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.

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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)

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

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
$\frac{3 \mathrm{R}}{4}, \frac{3 \mathrm{R}}{16}, \frac{7 \mathrm{R}}{144}$
D
$\frac{5 \mathrm{R}}{36}, \frac{3 \mathrm{R}}{16}, \frac{21 \mathrm{R}}{100}$

Did you get this right?

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.

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry