Q23 / 75 · JEE Main 2026 (24 January Shift 1) · Chemistry
The hydrogen spectrum consists of several spectral lines in Lyman series ($\mathrm{L}_{1}, \mathrm{~L}_{2}$, $\mathrm{L}_{3} \ldots ; \mathrm{L}_{1}$ has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series $\left(\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3} \ldots ; \mathrm{B}_{1}\right.$ has lowest energy among Balmer lines). The energy of $L_{1}$ is $x$ times the energy of $B_{1}$. The value of $x$ is $\_\_\_\_$ $\times 10^{-1}$ . (Nearest integer)
Correct answer: 54
Solution
The energy of a spectral line in the hydrogen spectrum is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ eV.
For the Lyman series, the lowest energy line $L_1$ corresponds to the transition from $n=2$ to $n=1$.
$E(L_1) = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4}$ eV.
For the Balmer series, the lowest energy line $B_1$ corresponds to the transition from $n=3$ to $n=2$.
$E(B_1) = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36}$ eV.
Given $E(L_1) = x \cdot E(B_1)$, we have:
$13.6 \times \frac{3}{4} = x \times 13.6 \times \frac{5}{36}$
$\frac{3}{4} = x \times \frac{5}{36}$
$x = \frac{3}{4} \times \frac{36}{5} = 3 \times \frac{9}{5} = \frac{27}{5} = 5.4$
The value of $x$ is $5.4 = 54 \times 10^{-1}$.
Comparing with the required format $x = \text{integer} \times 10^{-1}$, the integer is 54.
For the Lyman series, the lowest energy line $L_1$ corresponds to the transition from $n=2$ to $n=1$.
$E(L_1) = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4}$ eV.
For the Balmer series, the lowest energy line $B_1$ corresponds to the transition from $n=3$ to $n=2$.
$E(B_1) = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36}$ eV.
Given $E(L_1) = x \cdot E(B_1)$, we have:
$13.6 \times \frac{3}{4} = x \times 13.6 \times \frac{5}{36}$
$\frac{3}{4} = x \times \frac{5}{36}$
$x = \frac{3}{4} \times \frac{36}{5} = 3 \times \frac{9}{5} = \frac{27}{5} = 5.4$
The value of $x$ is $5.4 = 54 \times 10^{-1}$.
Comparing with the required format $x = \text{integer} \times 10^{-1}$, the integer is 54.
Hence, the correct answer is 54.