JEE Main 2026 — Chemistry Physical Chemistry

The energy of a spectral line in the hydrogen spectrum is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$E=13.6(n12​1​−n22​1​) eV.

For the Lyman series, the lowest energy line $L_1$L1​ corresponds to the transition from $n=2$n=2 to $n=1$n=1.

$E(L_1) = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4}$E(L1​)=13.6(121​−221​)=13.6(1−41​)=13.6×43​ eV.

For the Balmer series, the lowest energy line $B_1$B1​ corresponds to the transition from $n=3$n=3 to $n=2$n=2.

$E(B_1) = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36}$E(B1​)=13.6(221​−321​)=13.6(41​−91​)=13.6×365​ eV.

Given $E(L_1) = x \cdot E(B_1)$E(L1​)=x⋅E(B1​), we have:

$$13.6 \times \frac{3}{4} = x \times 13.6 \times \frac{5}{36}$$13.6×43​=x×13.6×365​

$$\frac{3}{4} = x \times \frac{5}{36}$$43​=x×365​

$$x = \frac{3}{4} \times \frac{36}{5} = 3 \times \frac{9}{5} = \frac{27}{5} = 5.4$$x=43​×536​=3×59​=527​=5.4

The value of $x$x is $5.4 = 54 \times 10^{-1}$5.4=54×10−1.

Comparing with the required format $x = \text{integer} \times 10^{-1}$x=integer×10−1, the integer is 54.