Final answer:
To find the wavelength of a photon when an electron drops from n=5 to n=2 in a hydrogen atom, we use the Rydberg formula and calculate that the wavelength is 434 nm.
Step-by-step explanation:
To calculate the wavelength of a photon emitted when an electron in a hydrogen atom transitions from the n=5 to the n=2 energy level, we use the Rydberg formula for hydrogen:
\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)
The Rydberg constant (R) for hydrogen is approximately \(1.097 \times 10^7 \, m^{-1}\). Substituting the given energy levels (n=2 and n=5) into the formula:
\( \frac{1}{\lambda} = 1.097 \times 10^7 \, m^{-1} \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \)
\( \frac{1}{\lambda} = 1.097 \times 10^7 \, m^{-1} \left( \frac{1}{4} - \frac{1}{25} \right) \)
\( \frac{1}{\lambda} = 1.097 \times 10^7 \, m^{-1} \times 0.21 \)
\( \frac{1}{\lambda} = 2.3037 \times 10^6 \, m^{-1} \)
So, \( \lambda = \frac{1}{2.3037 \times 10^6 \, m^{-1}}
= 4.34 \times 10^{-7} \, m \) or \( 434 \, nm \).
Converting meters to nanometers (1 m = 10^9 nm), we get:
\( \lambda = 434 \times 10^9 \, nm \)
Hence, the wavelength of the photon is 434 nm when rounded to three significant digits.