JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

In hydrogen atom spectrum, ( $R \to$ Rydberg's constant)

A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3 R}$

B. the Balmer series lies in the visible region of the spectrum

C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$

D. the minimum wavelength of Lyman series is $\frac{5}{4 R}$

Choose the correct answer from the options given below :

A
B, D Only
B
A, B and C Only
C
A, B and D Only
D
A, B Only

Solution

A. Lyman max wavelength corresponds to the smallest energy gap, $n = 2 \to n = 1$ : $\frac{1}{λ} = R (1 - \frac{1}{4}) = \frac{3 R}{4}$ , so $λ_{m a x} = \frac{4}{3 R}$ . ✓

B. The Balmer series ( $n \to 2$ ) lies in the visible region. ✓

C. Paschen min wavelength corresponds to $n = \infty \to n = 3$ : $\frac{1}{λ} = \frac{R}{9}$ , so $λ_{m i n} = \frac{9}{R}$ . ✓

D. Lyman min wavelength corresponds to $n = \infty \to n = 1$ : $\frac{1}{λ} = R$ , so $λ_{m i n} = \frac{1}{R}$ , not $\frac{5}{4 R}$ . ✗

$∴$ A, B and C only.

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