JEE Physics 2026 Question with Solution

Given: Statements about hydrogen atom spectrum and Rydberg's constant $R$.

Find: Which statements among A, B, C, D are correct.

Use the Rydberg formula:

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

Maximum wavelength corresponds to the smallest energy gap, so for a given series take $n_2 = n_1 + 1$. Minimum wavelength corresponds to the series limit, so take $n_2 = \infty$.

For Statement A: Lyman series has $n_1 = 1$. For maximum wavelength, take $n_2 = 2$.

$$\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$

Hence,

$$\lambda = \frac{4}{3R}$$

So A is correct.

For Statement B: Balmer series corresponds to $n_1 = 2$ and it lies in the visible region of the spectrum. So B is correct.

For Statement C: Paschen series has $n_1 = 3$. For minimum wavelength, take $n_2 = \infty$.

$$\frac{1}{\lambda} = R\left(\frac{1}{9} - 0\right) = \frac{R}{9}$$

Hence,

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

So C is correct.

For Statement D: Lyman series minimum wavelength occurs at $n_2 = \infty$.

$$\frac{1}{\lambda} = R(1 - 0) = R$$

Hence,

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

This is not $\frac{5}{4R}$, so D is incorrect.

Therefore, the correct statements are A, B and C only. The correct option is A.

Given: Hydrogen spectral series statements.

Find: The correct combination.

A quick way is to remember two facts:

1. Maximum wavelength in a series comes from the first line of that series.
2. Minimum wavelength comes from the series limit, that is $n_2 = \infty$.

• Lyman first line: $1 \to 2$ gives

$$\lambda = \frac{4}{3R}$$

So A is correct.

• Balmer series is known to lie in the visible region, so B is correct.

• Paschen limit: $n_1 = 3, n_2 = \infty$ gives

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

So C is correct.

• Lyman limit gives

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

not $\frac{5}{4R}$, so D is incorrect.

Therefore, the correct option is A.