Final answer:
The frequency of light emitted when an electron transitions from n = 3 to n = 2 in a hydrogen atom is approximately 4.57 x 10^14 Hz.
Step-by-step explanation:
The calculation of the frequency of light emitted from a hydrogen atom when an electron transitions from the n = 3 to the n = 2 energy level can be done using the Rydberg formula for hydrogen which relates the wavelength of the emitted light to the initial and final quantum numbers of the electron orbits:
1/λ = R_H (1/n_1^2 - 1/n_2^2)
where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 107 m-1), n_1 is the final quantum number (2), and n_2 is the initial quantum number (3).
Once we have the wavelength, we can find the frequency (f) by using the equation that relates the speed of light (c), frequency (f), and wavelength (λ):
f = c/λ
Here, c is the speed of light (approximately 3.00 x 108 m/s) and f is the frequency of the emitted light. So, we'll first calculate the wavelength using the Rydberg formula and then calculate the frequency using the speed of light.
For the transition from n = 3 to n = 2, this calculation results in a wavelength of 656 nm, which corresponds to red light within the Balmer series of hydrogen's emission spectrum. The frequency is then calculated as:
f = (3.00 x 108 m/s) / (656 x 10-9 m)
Doing the math we get:
f ≈ 4.57 x 1014 Hz
An electron falling from the n = 3 orbit to the n = 2 orbit emits light at a frequency of 4.57 x 1014 Hz.