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 $n_{3\to 2}$, 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,

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

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 ($\mathrm{\Delta }E$) of a particle in a 1D box that has undergone an energy level transition is,

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

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

$\mathrm{\Delta }E=\frac{h^2}{8m{L}^2}(n_f^2-n_i^2)$

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

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

If $n_f<n_i$, $\mathrm{\Delta }E$ 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 $n_{3\to 2}$ transition is,

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

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

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

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

Therefore,

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

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 $n_{3\to 2}$ transition of an electron in a 1D box with a length of 1.0 cm is,

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

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