Final answer:
Using the Rydberg formula, the wavelength of light emitted from the transition of an electron from n=4 to n=2 in a hydrogen atom is calculated to be approximately 486 nm, which is in the visible range of light.
Step-by-step explanation:
To calculate the wavelength of light emitted when an electron in a hydrogen atom undergoes a transition from n=4 to n=2, we can use the Rydberg formula for hydrogen:
\[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]
Where:
- \(\lambda\) is the wavelength of the emitted light
- \(R_H\) is the Rydberg constant (\(1.097 \times 10^7 m^{-1}\))
- \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final energy levels, respectively
Plugging the values into the formula:
\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{1}{2^2} - \frac{1}{4^2}\right)\]
\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{1}{4} - \frac{1}{16}\right)\]
\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{4}{16} - \frac{1}{16}\right)\]
\[\frac{1}{\lambda} = (1.097 \times 10^7 m^{-1}) \left(\frac{3}{16}\right)\]
\[\frac{1}{\lambda} = 2.055 \times 10^6 m^{-1}\]
\[\lambda = \frac{1}{2.055 \times 10^6 m^{-1}}\]
\[\lambda \approx 4.86 \times 10^{-7} m\]
The wavelength of light emitted by this transition is approximately 486 nm, which falls within the visible light spectrum, specifically within the blue range.