JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

The hydrogen spectrum consists of several spectral lines in Lyman series ( $L_{1}, L_{2}, L_{3}, \dots$ ; $L_{1}$ has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series ( $B_{1}, B_{2}, B_{3}, \dots$ ; $B_{1}$ has lowest energy among Balmer lines). The energy of $L_{1}$ is $x$ times the energy of $B_{1}$ . The value of $x$ is ______ $\times 1 0^{- 1}$ (Nearest integer)

Your answer

Solution

$L_{1}$ corresponds to the lowest-energy Lyman transition: $n = 2 \to n = 1$ .

$B_{1}$ corresponds to the lowest-energy Balmer transition: $n = 3 \to n = 2$ .

Using $Δ E = 13.6 Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ eV for H ( $Z = 1$ ):

$Δ E (L_{1}) = 13.6 (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4}$

$Δ E (B_{1}) = 13.6 (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36}$

$\frac{Δ E ( L _{1} )}{Δ E ( B _{1} )} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$ .

$x = 5.4 = 54 \times 1 0^{- 1}$ .

Answer: $54$ .

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