JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly _____ nm.

A
1875
B
1550
C
1217
D
1784

Solution

Smallest Lyman wavelength corresponds to $n_{1} = 1, n_{2} = \infty$ : $\frac{1}{λ} = R (1) \Rightarrow R = \frac{1}{91} n m^{- 1}$ .

Largest Balmer wavelength is the $n = 3 \to 2$ transition: $\frac{1}{λ _{B}} = R (\frac{1}{4} - \frac{1}{9}) = \frac{R \cdot 5}{36}$ , so $λ_{B} = \frac{36 \cdot 91}{5} \approx 655.2$ nm.

Largest Paschen wavelength is the $n = 4 \to 3$ transition: $\frac{1}{λ _{P}} = R (\frac{1}{9} - \frac{1}{16}) = \frac{7 R}{144}$ , so $λ_{P} = \frac{144 \cdot 91}{7} \approx 1872$ nm.

Difference $\approx 1872 - 655.2 \approx 1217$ nm.

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