# Quantum Chemistry/Example 4

## Example 4

Write a question about calculating the frequency of a photon to calculate the energy to transition between two levels of an electron in a 1D box

## 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 ${\displaystyle n_{{3}\rightarrow {2}}}$ , 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 (${\displaystyle {E_{n}}}$ ) is,

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

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

Since ${\displaystyle {E_{n}}\propto {n^{2}}}$  (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 (${\displaystyle \Delta {E}}$ ) of a particle in a 1D box that has undergone an energy level transition is,

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

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

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

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

If ${\displaystyle n_{f}>n_{i}}$ , ${\displaystyle \Delta {E}}$  is a positive value; photon absorbed.

If ${\displaystyle n_{f} , ${\displaystyle \Delta {E}}$  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 ${\displaystyle n_{{3}\rightarrow {2}}}$  transition is,

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

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

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

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

Therefore,

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

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,

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

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

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

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

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

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

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

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