Final answer:
The wavelength of light emitted during a transition from n = 6 to n = 3 in a hydrogen atom is approximately 820 nm, which is in the infrared spectrum.
Step-by-step explanation:
The wavelength of light emitted when an electron in a hydrogen atom transitions from an energy level n = 6 to n = 3 can be calculated using the Rydberg formula for hydrogen atom transitions:
Rydberg formula: \(\frac{1}{\lambda} = R \left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)\)
Where:
\(\lambda\) is the wavelength of the light emitted,\(R\) is the Rydberg constant (1.097 x 107 m-1),\(n_{1}\) and \(n_{2}\) are the initial and final energy levels of the electron (6 and 3, respectively).
Plugging in the values we get:
\(\frac{1}{\lambda} = 1.097 \times 107 \left(\frac{1}{3^{2}} - \frac{1}{6^{2}}\right)\)
\(\frac{1}{\lambda} = 1.097 \times 107 \left(\frac{1}{9} - \frac{1}{36}\right)\)
\(\frac{1}{\lambda} = 1.097 \times 107 \times \frac{4}{36}\)
\(\frac{1}{\lambda} = 1.097 \times 107 \times \frac{1}{9}\)
\(\lambda = \frac{9}{1.097 \times 107}\)
After calculating, we find that \(\lambda\approx 820 nm\), which falls in the infrared region of the electromagnetic spectrum.