Correct Answer: 54

Explanation

For hydrogen atom, energy of emitted photon is given by

\[ E = 13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \text{eV} \]

Energy of L₁ (Lyman series):

For Lyman series, transitions end at \(n_1 = 1\).

For L₁, the transition is from \(n_2 = 2 \to n_1 = 1\).

\[ E_{L_1} = 13.6\left(1-\frac{1}{4}\right) \]

\[ E_{L_1} = 13.6 \times \frac{3}{4} \]

\[ E_{L_1} = 10.2\ \text{eV} \]

Energy of B₁ (Balmer series):

For Balmer series, transitions end at \(n_1 = 2\).

For B₁, the transition is from \(n_2 = 3 \to n_1 = 2\).

\[ E_{B_1} = 13.6\left(\frac{1}{4}-\frac{1}{9}\right) \]

\[ E_{B_1} = 13.6 \times \frac{5}{36} \]

\[ E_{B_1} \approx 1.89\ \text{eV} \]

Ratio of energies:

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

\[ x = 54 \times 10^{-1} \]

Hence, the required value of x = 54.

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