JEE Chemistry 2026 Question with Solution

Given:

• Hydrogen spectral lines in the Lyman series and Balmer series
• $L_1$ is the transition from $n_i = 2$ to $n_f = 1$
• $B_1$ is the transition from $n_i = 3$ to $n_f = 2$

Find: The value to be filled in the blank for $x \times 10^{-1}$ where $x = \dfrac{E_{L_1}}{E_{B_1}}$

For hydrogen, the transition energy is

$$E = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \, \text{eV}$$

For $L_1$:

$$E_{L_1} = 13.6\left(1 - \frac{1}{4}\right)$$ $$E_{L_1} = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV}$$

For $B_1$:

$$E_{B_1} = 13.6\left(\frac{1}{4} - \frac{1}{9}\right)$$ $$E_{B_1} = 13.6 \times \frac{5}{36} \approx 1.89 \, \text{eV}$$

Now,

$$x = \frac{E_{L_1}}{E_{B_1}} = \frac{10.2}{1.89} \approx 5.4$$

So,

$$x = 54 \times 10^{-1}$$

Therefore, the nearest integer to be filled in the blank is 54.

Given: $L_1$ corresponds to $2 \to 1$ and $B_1$ corresponds to $3 \to 2$.

Find: The integer in _____ \times 10^{-1}.

Using the hydrogen energy-gap formula:

$$E_{L_1} = 13.6\left(1 - \frac{1}{4}\right) = 13.6 \cdot \frac{3}{4}$$ $$E_{B_1} = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \cdot \frac{5}{36}$$

Hence,

$$x = \frac{E_{L_1}}{E_{B_1}} = \frac{\frac{3}{4}}{\frac{5}{36}} = \frac{3}{4} \cdot \frac{36}{5} = \frac{27}{5} = 5.4$$

Rewriting in the asked form,

$$5.4 = 54 \times 10^{-1}$$

Therefore, the required nearest integer is 54.