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

JEE Main2026Jan 22, Shift 2Chemistry

The energy of first (lowest) Balmer line of H atom is $x$ J. The energy (in J) of second Balmer line of H atom is :

Solution

Balmer series: $n_{1} = 2$ . The first Balmer line is the transition $n = 3 \to 2$ ; the second is $n = 4 \to 2$ .

$Δ E \propto (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ .

For first line: $\frac{1}{4} - \frac{1}{9} = \frac{5}{36}$ . For second line: $\frac{1}{4} - \frac{1}{16} = \frac{3}{16}$ .

Ratio $= \frac{3/16}{5/36} = \frac{3 \times 36}{16 \times 5} = \frac{27}{20} = 1.35$ .

Hence second-line energy $= 1.35 x$ .

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.