Final answer:
The wavelength of the photon emitted during a transition from a 4d to a 2p orbital in a hydrogen atom is calculated using the Rydberg formula, giving us a result of approximately 486.2 nm, which is option B).
Step-by-step explanation:
To calculate the wavelength of the photon emitted when an electron in a hydrogen atom drops from a 4d orbital to a 2p orbital, we use the Rydberg formula. The Rydberg formula is given by:
\(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
Where \(\lambda\) is the wavelength of the emitted photon, \(R_H\) is the Rydberg constant, \(n_1\) is the principal quantum number for the lower energy level, and \(n_2\) is the principal quantum number for the higher energy level. For a transition from 4d to 2p:
- \(n_1 = 2\) (for 2p)
- \(n_2 = 4\) (for 4d)
- \(R_H = 1.097 \times 10^7 m^{-1}\)
Plugging in the values, we get:
\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)\)
\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)\)
\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{4}{16} - \frac{1}{16} \right)\)
\(\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{16}\)
\(\frac{1}{\lambda} = 2.053125 \times 10^6 m^{-1}\)
\(\lambda = \frac{1}{2.053125 \times 10^6}\)
\(\lambda \approx 486.2 nm\)
The correct answer is B) 486.2 nm. This wavelength corresponds to light in the visible spectrum, specifically to the blue light range.