Question

Energy of first Balmer line of H-atom is \( x \) kJ. The energy of the second Balmer line of H-atom is _____

• A. \( x \)
• B. 1.35 \( x \)
• C. 2 \( x \)
• D. \( x / 1.35 \)

Hint:

The energy of the lines in the Balmer series follows a pattern that can be calculated using the Rydberg formula, with each successive line having a higher energy.

Answer 1

The Correct Option is B

The question involves understanding the energy levels of hydrogen atoms, particularly the Balmer series, which pertains to transitions where the final energy level is \(n=2\). To solve this, we relate the energies of different lines in the Balmer series using the Rydberg formula:

• The Rydberg formula for wavelengths in the Balmer series is given by:

\(\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)\), where \(R_H\) is the Rydberg constant and \(n > 2\).

• For the first Balmer line, \(n=3\), the change in energy is:

\(\Delta E_1 = R_H \cdot h \cdot c \left( \frac{1}{4} - \frac{1}{9} \right)\)

Here, \( x \) kJ represents this energy.

• For the second Balmer line, \(n=4\), the change in energy is:

\(\Delta E_2 = R_H \cdot h \cdot c \left( \frac{1}{4} - \frac{1}{16} \right)\)

• The ratio of the energies of the second Balmer line to the first is given by:

\(\frac{\Delta E_2}{\Delta E_1} = \frac{\frac{1}{4} - \frac{1}{16}}{\frac{1}{4} - \frac{1}{9}}\)

Calculating this gives:

\(\frac{12}{9} \times \frac{3}{1} = \frac{3}{4} \times \frac{16}{12} = 1.35\)

• Therefore, the energy of the second Balmer line is 1.35 times the energy of the first Balmer line.

Hence, the energy of the second Balmer line is \(1.35 \, x\) kJ.

• Thus, the correct answer is the option:

1.35 x