Question

The wavelength of the first line in the Balmer series of hydrogen spectrum is 656 nm. What is the wavelength of the second line?

• A. \(486\, \text{nm}\)
• B. \(434\, \text{nm}\)
• C. \(397\, \text{nm}\)
• D. \(410\, \text{nm}\)

Hint:

In the Balmer series, the first few lines correspond to transitions to \(n = 2\) from \(n = 3, 4, 5, \ldots\). Use the Rydberg formula to find successive wavelengths.

Answer 1

The Correct Option is A

Step 1: Recall the Balmer series formula.
The Balmer series corresponds to transitions from \(n \geq 3\) to \(n = 2\). The wavelength is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \] where \(R_H = 1.097 \times 10^7\, \text{m}^{-1}\)
Step 2: First line in Balmer series.
This corresponds to \(n = 3 \to 2\): \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right) \]
Step 3: Second line in Balmer series.
This corresponds to \(n = 4 \to 2\): \[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right) \] \[ \Rightarrow \frac{\lambda_2}{\lambda_1} = \frac{5/36}{3/16} = \frac{5 \cdot 16}{36 \cdot 3} = \frac{80}{108} = \frac{20}{27} \Rightarrow \lambda_2 = \frac{20}{27} \cdot 656 \approx 486\, \text{nm} \]