JEE Chemistry 2026 Question with Solution

Given: The wave numbers of three spectral lines are listed, and we must identify the set belonging to the Balmer series.

Find: Which option contains only Balmer series wave numbers.

Concept: The wave number of a hydrogen spectral line is given by

$$\bar{\nu} = R\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \quad n_2>n_1$$

For the Balmer series,

$$n_1 = 2,\quad n_2 = 3,4,5,\ldots$$

So the Balmer series wave numbers must have the form

$$\bar{\nu}=R\!\left(\frac{1}{4}-\frac{1}{n^2}\right)$$

Step 1: Check each expression in option D.

For $\dfrac{5R}{36}$,

$$\frac{5}{36}=\frac{1}{4}-\frac{1}{9}$$

Hence, $n_2=3$.

For $\dfrac{3R}{16}$,

$$\frac{3}{16}=\frac{1}{4}-\frac{1}{16}$$

Hence, $n_2=4$.

For $\dfrac{21R}{100}$,

$$\frac{21}{100}=\frac{1}{4}-\frac{1}{25}$$

Hence, $n_2=5$.

All three expressions correspond to transitions ending at $n=2$.

Therefore, the correct option is D.

Given: Balmer series lines of hydrogen.

Find: A quick way to identify the correct set.

For the Balmer series, always factor out

$$\frac{1}{4}$$

and check whether the remaining term is of the form

$$\frac{1}{n^2}$$

so that

$$\bar{\nu}=R\left(\frac{1}{4}-\frac{1}{n^2}\right)$$

In option D:

$$\frac{5}{36}=\frac{1}{4}-\frac{1}{9}, \quad \frac{3}{16}=\frac{1}{4}-\frac{1}{16}, \quad \frac{21}{100}=\frac{1}{4}-\frac{1}{25}$$

These correspond to $n=3,4,5$ respectively, so all are Balmer lines.

Therefore, the correct option is D.