JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Modern Physics

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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✓ Correct
(D)1784

Answer: (C)

Solution

The shortest Lyman wavelength is the series limit

\infty \to 1

\frac{1}{λ} = R

, so

R = \frac{1}{91} n m^{- 1}

. The longest Balmer line is

3 \to 2

\frac{1}{λ _{B}} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}

, giving

λ_{B} = \frac{36}{5 R} = \frac{36 \times 91}{5} \approx 655 nm

. The longest Paschen line is

4 \to 3

\frac{1}{λ _{P}} = R (\frac{1}{9} - \frac{1}{16}) = \frac{7 R}{144}

, giving

λ_{P} = \frac{144}{7 R} = \frac{144 \times 91}{7} = 1872 nm

. The difference is

λ_{P} - λ_{B} \approx 1872 - 655 = 1217 nm

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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