Using the Balmer formula for hydrogen, what is the wavelength (in nm) of the spectral line produced when an electron transitions from n = 8 to n = 2?

• A. 434 nm
• B. 388 nm
• C. 656 nm
• D. 486 nm

Answer: (B)

Verified step by step guidance

1. Identify the Balmer formula for the wavelength of light emitted during an electron transition in a hydrogen atom: \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\), where \(\lambda\) is the wavelength, \(R\) is the Rydberg constant (\$1.097 \times 10^7 \ \text{m}^{-1}\(), \)n_1\( is the lower energy level, and \)n_2$ is the higher energy level.
2. Assign the given values: the electron transitions from \(n_2 = 8\) to \(n_1 = 2\) (since the electron moves to a lower energy level, \(n_1 < n_2\)).
3. Substitute the values into the formula: \(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{8^2} \right)\).
4. Calculate the difference inside the parentheses: \(\frac{1}{4} - \frac{1}{64}\), which simplifies the expression for \(\frac{1}{\lambda}\).
5. Take the reciprocal of \(\frac{1}{\lambda}\) to find \(\lambda\) in meters, then convert the wavelength to nanometers by multiplying by \$10^9\( (since \)1 \ \text{m} = 10^9 \ \text{nm}$).