JEE Chemistry 2025 Question with Solution

Given: wavelength is about $900 \, \text{nm}$.

Find: the suitable spectral line of the hydrogen atom.

Use the Rydberg relation:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

The solution identifies the relevant hydrogen series by comparing the given wavelength with the spectral regions.

Given:

$$\lambda = 900 \, \text{nm} = 900 \times 10^{-9} \, \text{m} = 9000 \, \text{\AA}$$

and

$$R_H = 10^5 \, \text{cm}^{-1} = 10^7 \, \text{m}^{-1}$$

Substituting:

$$\frac{1}{9000 \times 10^{-10}} = 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

So,

$$\frac{1}{9000} = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

Now compare the series:

• Lyman series: $n_1 = 1$, ultraviolet region
• Balmer series: $n_1 = 2$, visible region
• Paschen series: $n_1 = 3$, infrared region

Since $900 \, \text{nm}$ lies in the infrared region, the suitable spectral line belongs to the Paschen series. The solution concludes that the matching line is $\infty \to 3$.

Therefore, the correct option is A, that is Paschen series, $\infty \to 3$.

Given: hydrogen atom transition associated with wavelength $900 \, \text{nm}$.

Find: which listed transition gives this wavelength.

For the Paschen limit, take the transition $\infty \to 3$. Then

$$\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)$$

Using the values stated in the solution:

$$\frac{1}{\lambda} = \frac{R}{9}$$

This gives a wavelength near $900 \, \text{nm}$.

Hence, the suitable line is Paschen series, $\infty \to 3$.

Therefore, the correct option is A.