Answer:
To calculate the energy, frequency, and wavelength of the emitted photon when an electron drops from the n=5 to n=2 energy level, we can use the following equations:
ΔE = E_final - E_initial = -(R_h / n_final^2) + (R_h / n_initial^2)
E = h * f
c = λ * f
where:
ΔE is the change in energy
E_final is the energy of the final level
E_initial is the energy of the initial level
R_h is the Rydberg constant (1.0974 x 10^7 m^-1)
n_final is the final level
n_initial is the initial level
h is Planck's constant (6.626 x 10^-34 J s)
f is the frequency of the photon
c is the speed of light (2.998 x 10^8 m/s)
λ is the wavelength of the photon
Plugging in the values:
ΔE = -(1.0974 x 10^7 m^-1) * [1/2^2 - 1/5^2] = -4.0948 x 10^-19 J
E = h * f
f = E / h = (-4.0948 x 10^-19 J) / (6.626 x 10^-34 J s) = 6.1707 x 10^14 Hz
c = λ * f
λ = c / f = (2.998 x 10^8 m/s) / (6.1707 x 10^14 Hz) = 4.859 x 10^-7 m = 485.9 nm
Therefore, the energy of the emitted photon is -4.0948 x 10^-19 J, the frequency is 6.1707 x 10^14 Hz, and the wavelength is 485.9 nm.