Question

An electron in a harmonic oscillator potential emits a photon with a wavelength of \(360 \mathrm{nm}\) as it undergoes a \(3 \rightarrow 1\) quantum jump. What is the wavelength of the photon emitted in a \(3 \rightarrow 2\) quantum jump? (Hint: The energy of the photon is equal to the energy difference between the initial and the final state of the electron.)

Short answer

Answer: The wavelength of the photon emitted in a \(3 \rightarrow 2\) quantum jump is \(540 \mathrm{nm}\).

Step-by-step solution

Calculate the energy difference for the 3 → 1 jump

First, we need to calculate the energy difference between the initial (n=3) and final (n=1) states. We can find it by using the energy of the emitted photon, using the following formula: \(E_{photon} = hf\) Where \(E_{photon}\) is the energy of the photon, \(h\) is the Planck's constant, and \(f\) is the frequency of the photon. Since we are given the wavelength, we can calculate the frequency using the speed of light (\(c\)): \(f = \frac{c}{\lambda}\) Therefore: \(E_{photon} = h\frac{c}{\lambda}\) Note that the energy difference between the n=3 and n=1 state is equal to the energy of the emitted photon: \(\Delta E_{31} = E_{photon}\)

Calculate the energy difference for the 3 → 2 jump

Next, we need to calculate the energy difference between the initial (n=3) and final (n=2) states. Since the harmonic oscillator has quantized energy levels proportional to the quantum number, we can express the energy difference as follows: \(\Delta E_{32} = \frac{2}{3} \Delta E_{31}\) Using the energy difference calculated in step 1, we can find the energy difference for the \(3 \rightarrow 2\) jump: \(\Delta E_{32} = \frac{2}{3} E_{photon}\)

Calculate the wavelength of the photon emitted in the 3 → 2 jump

Now we can find the wavelength of the photon emitted in the \(3 \rightarrow 2\) jump using the energy difference calculated in step 2 and the same formula as in step 1: \(E_{photon_{32}} = h\frac{c}{\lambda_{32}}\) Since the energy difference for the \(3 \rightarrow 2\) jump is equal to the energy of the emitted photon in this case: \(\Delta E_{32} = E_{photon_{32}}\) Therefore: \(\lambda_{32} = h\frac{c}{\Delta E_{32}}\) Now substitute the value of \(\Delta E_{32}\) found in step 2: \(\lambda_{32} = h\frac{c}{\frac{2}{3}E_{photon}}\) Finally, substitute the value of the given wavelength (\(360 \mathrm{nm}\)) and solve for \(\lambda_{32}\): \(\lambda_{32} = \frac{3}{2}\lambda\) \(\lambda_{32} = \frac{3}{2}(360 \mathrm{nm})\) \(\lambda_{32} = 540 \mathrm{nm}\) So, the wavelength of the photon emitted in a \(3 \rightarrow 2\) quantum jump is \(540 \mathrm{nm}\).