Quantum Chemistry/Example 4

Question
An electron in a 1D box emits a photon as the electron transitions to a lower energy level. If the length of the 1D box is equal to 1.0 cm, and the quantum number transition is $$n_{{3}\rightarrow{2}}$$, what is the electromagnetic radiation frequency of the emitted photon?

Solution:

The energy level of a particle in a 1D box at a specific quantum number ($${E_n}$$) is,

$$E_n = \frac{h^2} {8mL^2} {n^2}$$

Where $$h$$ is equal to Planck's constant (6.62607015 x 10-34 J$$\cdot$$s), $$n$$ is equal to the quantum number ($$n$$ = 1, 2, 3, ...), $$m$$ is equal to the mass of the particle, and $$L$$ is equal to the length of the 1D box. For an electron, the mass is equal to 9.10938356 x 10-31 kg.

Since $${E_n}\propto{n^2}$$ (assuming the mass and box length are constant), the energy level increases by a factor of 4 as the quantum number increases by a factor of 2. Therefore, if a particle in a 1D box undergoes an energy level transition, there is a difference between the initial and final quantum number energy levels. The energy level difference ($$\Delta {E}$$) of a particle in a 1D box that has undergone an energy level transition is,

$$\Delta {E} = {E_{n_f}} - {E_{n_i}}$$

$$= \left[\frac{h^2} {8mL^2} {n^2_f}\right] - \left[\frac{h^2} {8mL^2} {n^2_i}\right]$$

$$\Delta {E} = \frac{h^2} {8mL^2} {({n^2_f} - {n^2_i}})$$

Where $$n_f$$ is equal to the final quantum number, and $$n_i$$ is equal to the initial quantum number.

If $$n_f > n_i$$, $$\Delta {E}$$ is a positive value; photon absorbed.

If $$n_f < n_i$$, $$\Delta {E}$$ is a negative value; photon emitted.

Therefore, the energy level difference of an electron in a 1D box with a length of 1.0 cm, which has undergone a $$n_{{3}\rightarrow{2}}$$ transition is,

$$\Delta {E} = \left[\frac{(6.6261 \times 10^{-34} \text{ J}\text{ s})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({2}^2 - {3}^2}) = \left[\frac{(6.6261 \times 10^{-34} \text{ m}^2\text{ kg}\text{ s}^{-1})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({2}^2 - {3}^2})$$

$$\Delta {E} = -3.01 \times 10^{-33} \text{ J}$$

The energy level difference for the electron which underwent a $$n_{{3}\rightarrow{2}}$$ transition is equal to -3.01 × 10-33 J. Since $$n_f < n_i$$, 3.01 × 10-33 J was emitted from the electron. If the electron underwent a $$n_{{2}\rightarrow{3}}$$ transition, the electron would absorb the same amount of energy that was emitted from the $$n_{{3}\rightarrow{2}}$$ transition which was 3.01 × 10-33 J.

$$\Delta {E} = \left[\frac{(6.6261 \times 10^{-34} \text{ m}^2\text{ kg}\text{ s}^{-1})^2} {8(9.1094 \times 10^{-31}\text{ kg})(1.0 \times 10^{-2}\text{ m})^2}\right] {({3}^2 - {2}^2}) = 3.01 \times 10^{-33} \text{ J}$$

Therefore,

$$E_{\text{photon}_{n_{2\rightarrow3}}} = E_{\text{photon}_{n_{3\rightarrow2}}}$$

The energy of a photon has a specific frequency of electromagnetic (EM) radiation, and the energy is directly proportional to the frequency. The energy of the photon is equal to,

$$E = h\cdot{f}$$

Where $$h$$ is equal to Planck's constant (6.62607015 x 10-34 J$$\cdot$$s), and $$f$$ is equal to EM radiation frequency.

Rearranging this equation allows for the calculation of the photon EM radiation frequency,

$$E = h\cdot{f}$$

$$f = \frac{E} {h}$$

The calculated photon energy was equal to 3.01 × 10-33 J, therefore the EM radiation frequency of the emitted photon from the $$n_{{3}\rightarrow{2}}$$ transition of an electron in a 1D box with a length of 1.0 cm is,

$$f = \frac{3.01 \times 10^{-33} \text{ J}} {6.6261 \times 10^{-34} \text{ J}\text{ s}}$$

$$f = 4.54 \text{s}^{-1} = 4.54 \text{Hz}$$