JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Atomic Structure

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

Answer: (B)

Solution

The first (lowest-energy) Balmer line is the

n = 3 \to 2

transition:

\overset{ν}{ˉ}_{1} = R_{H} (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R _{H}}{36}

. The first Brackett line is the

n = 5 \to 4

transition:

\overset{ν}{ˉ}_{2} = R_{H} (\frac{1}{4 ^{2}} - \frac{1}{5 ^{2}}) = \frac{9 R _{H}}{400}

. The ratio is

\frac{ν ˉ _{1}}{ν ˉ _{2}} = \frac{5/36}{9/400} = \frac{2000}{324} = \frac{500}{81}

. Written with the first term as 5 this is

5 : \frac{81 \times 5}{500} = 5 : 0.81

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option