Question: An excited hydrogen atom undergoes a transition from energy level n=4 to energy level n=2.
a) Calculate the wavelength of the photon emitted during this transition.
b) Calculate the frequency of the emitted photon.
c) Identify the series to which this transition belongs and explain.
Answer:
a) To calculate the wavelength of the photon emitted during this transition, we can use the Rydberg formula given by:
1/λ = R_H * (1/n_f^2 - 1/n_i^2)
Where: λ is the wavelength of the photon emitted R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m^-1) n_f is the final energy level (n=2 in this case) n_i is the initial energy level (n=4 in this case)
Plugging in the values into the formula, we get:
1/λ = (1.097 × 10^7 m^-1) * (1/2^2 - 1/4^2)
Simplifying the equation:
1/λ = (1.097 × 10^7 m^-1) * (1/4 - 1/16)
1/λ = (1.097 × 10^7 m^-1) * (3/16)
1/λ = 0.20585 × 10^7 m^-1
Taking the reciprocal, we have:
λ = 1/(0.20585 × 10^7 m^-1)
λ ≈ 4.85 × 10^-8 m
Therefore, the wavelength of the emitted photon is approximately 4.85 × 10^-8 meters.
b) To calculate the frequency of the emitted photon, we can use the equation:
c = λ * ν
Where: c is the speed of light (approximately 3 × 10^8 m/s) λ is the wavelength of the photon emitted (4.85 × 10^-8 m)
Plugging in the values, we have:
3 × 10^8 m/s = (4.85 × 10^-8 m) * ν
Simplifying the equation, we can solve for ν:
ν = (3 × 10^8 m/s) / (4.85 × 10^-8 m)
ν ≈ 6.18 × 10^15 Hz
Therefore, the frequency of the emitted photon is approximately 6.18 × 10^15 Hz.
c) This transition belongs to the Balmer series. The Balmer series corresponds to transitions where the final energy level (n_f) is 2. The series is characterized by the emission of visible light. In this case, since the transition is from n=4 to n=2, the emitted photon falls in the visible light region of the electromagnetic spectrum.
The Balmer series is particularly important because it is the visible light emitted from hydrogen that helped confirm the existence of quantized energy levels and the Bohr model of the atom.