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,

$\Delta {E}={E_{n_{f}}}-{E_{n_{i}}}$

$=\left[{\frac {h^{2}}{8mL^{2}}}{n_{f}^{2}}\right]-\left[{\frac {h^{2}}{8mL^{2}}}{n_{i}^{2}}\right]$

$\Delta {E}={\frac {h^{2}}{8mL^{2}}}{({n_{f}^{2}}-{n_{i}^{2}}})$

Where $n_{f}$ is equal to the final quantum number, and $n_{i}$ is equal to the initial quantum number.

If $n_{f}>n_{i}$, $\Delta {E}$ 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,

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

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

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

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