JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Modern Physics

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Answer: (B)

Solution

In the Lyman series every transition ends at

n = 1

, and the largest wavelength corresponds to the smallest energy jump,

n = 2 \to n = 1

\frac{1}{λ _{L}} = R (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = \frac{3 R}{4} .

In the Balmer series every transition ends at

n = 2

, and the largest wavelength corresponds to

n = 3 \to n = 2

\frac{1}{λ _{B}} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R}{36} .

The ratio of the two largest wavelengths is therefore

\frac{λ _{L}}{λ _{B}} = \frac{4/3 R}{36/5 R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} .

The largest Lyman wavelength to the largest Balmer wavelength stands in the ratio

5 : 27

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option