The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is

Q. The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :

A. \( \dfrac{x}{1.35} \)

B. \( 1.35x \)

C. \( x^2 \)

D. \( 2x \)

Correct Answer: 1.35x

Explanation

Balmer series corresponds to electronic transitions from higher energy levels to \( n = 2 \).

Energy of emitted photon is given by:

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

For the first (lowest) Balmer line:

Transition: \( n = 3 \rightarrow n = 2 \)

\[ E_1 = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} \]

This energy is given as \( x \) J.

For the second Balmer line:

Transition: \( n = 4 \rightarrow n = 2 \)

\[ E_2 = 13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} \]

Ratio of energies:

\[ \frac{E_2}{E_1} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3 \times 36}{5 \times 16} = \frac{108}{80} = 1.35 \]

\[ E_2 = 1.35 \, x \]

Hence, the correct answer is 1.35x.