An electron de-excites from the fourth quantum level, to the third and then directly to the ground state. Two photons are emitted. How does the sum of their frequencies compare with the frequency of the single photon that would be emitted by de-excitation from the fourth level directly to the ground state?
To understand how the sum of the frequencies of the two photons compares with the frequency of a single photon, let's break it down step by step:
First, let's consider the energy levels associated with each transition:
- From the fourth quantum level to the third quantum level
- From the third quantum level to the ground state (first level)
In an electron de-excitation, the energy difference between the initial and final quantum levels corresponds to the energy of the emitted photon. The energy of a photon is given by the equation E = hf, where E is energy, h is Planck's constant (6.626 x 10^-34 J·s), and f is frequency.
Now, let's determine the frequency of the photons emitted in each transition:
1. Transition from the fourth to the third level:
The energy difference ΔE1 between the fourth and third levels will be equal to the energy of the first photon emitted. Let's denote the frequency of this photon as f1.
So, f1 = ΔE1 / h
2. Transition from the third level to the ground state:
Similarly, the energy difference ΔE2 between the third level and the ground state will be equal to the energy of the second photon emitted. Let's denote the frequency of this photon as f2.
So, f2 = ΔE2 / h
Now, let's consider the original question regarding the sum of the frequencies of these two photons compared to the frequency of a single photon emitted directly from the fourth level to the ground state (let's call this frequency f3):
The sum of the frequencies of the two photons (f1 + f2) will be equal to the frequency of a single photon emitted directly from the fourth level to the ground state (f3).
Therefore, f1 + f2 = f3.
In this case, since the electron de-excites directly from the fourth level to the ground state, we can assume f1 = f3. This is because the energy difference between these two levels will be the same, regardless of whether the transition occurs in one step or two steps.
So, in conclusion, the sum of the frequencies of the two photons emitted during the two-step de-excitation process will be equal to the frequency of a single photon emitted directly from the fourth level to the ground state.
Energy is the same as frequency.
E1+ E2 = E3
F1 + F2 = F3