In the Balmer series of hydrogen, what is the wavelength (in nm) of the spectral line corresponding to the electron transition from n = 4 to n = 2?

• A. 486 nm
• B. 434 nm
• C. 410 nm
• D. 656 nm

Answer: (A)

Verified step by step guidance

1. Identify the initial and final energy levels for the electron transition. Here, the electron moves from the higher energy level \(n_i = 4\) to the lower energy level \(n_f = 2\) in the Balmer series.
2. Recall the Rydberg formula for the wavelength of emitted or absorbed light during an electron transition in a hydrogen atom: \[\lambda = \frac{1}{R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)}\] where \(\lambda\) is the wavelength, \(R\) is the Rydberg constant (\$1.097 \times 10^7 \ \text{m}^{-1}\(), \)n_f\( is the final energy level, and \)n_i$ is the initial energy level.
3. Substitute the known values into the formula: \[\lambda = \frac{1}{1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)}\]
4. Calculate the difference inside the parentheses: \[\frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16}\]
5. After finding \(\lambda\) in meters, convert it to nanometers by multiplying by \$10^9\( (since \)1 \ \text{m} = 10^9 \ \text{nm}$). This will give the wavelength of the spectral line in nanometers.