Quantum Chemistry/Example 30

Energy in the particle in a 1D box model
Energy in quantum mechanical systems is quantized, unlike in classically mechanical systems, where the energy can take any value. The energy of a quantum mechanical system may only hold specific discrete values, which are known as s. Consider the model for the 1D. In this model, the possible discrete energy levels are given by the relation,

$$ The ground state of any quantum mechanical system is the state with the lowest possible energy (otherwise known as the of the system). Defined in the equation for the energy levels of a particle in a 1D box, the quantum number $$n$$ can hold any integer value equal to or greater than one and as such, the ground state of this model occurs when $$n=1$$. All energy levels above the ground state are referred to as excited states. There are an infinite number of these states and for the particle in a 1D box, are defined by the quantum number $$n$$, like the ground state. The first excited state is the energy level one above the ground state, in which $$n=2$$, the second excited state is the energy level one above the first excited state, where $$n=3$$, and so forth.

Particle transitions between states
A particle in a 1D box can transition between energy levels by the means of three different processes, whose probability can be measured by. The transitional processes involve electromagnetic radiation in the form of photons. They are the following:

The energy of a photon can be determined by the relationship between a photon's frequency of oscillation and energy, known as the , $$ In quantum chemistry, it is often more convenient to express frequency in terms of wavenumber ($$\tilde{\nu}$$), which has units of $$\text{cm}^{-1}$$ compared to $$\text{s}^{-1}$$. There is a simple conversion as frequency and wavenumber are related to the speed of light by,
 * Spontaneous emission: The spontaneous decay of a particle from a higher energy state to a lower energy state, emitting a photon of the transition energy.
 * Stimulated emission: The decay of a particle from a higher energy state to a lower energy state stimulated by the presence of a photon of the same energy as the transition.
 * Photon absorption: The promotion of a particle from a lower energy state to a higher energy state caused by the absorption of a photon of the transition energy.
 * $$\tilde{\nu}=\frac{\nu}{c} \, \text{, where} \ c \ \text{is the speed of light in cm/s.}$$

As such, the Planck relation can be rewritten as, $$ This relation allows the calculation of the energy of a photon involved in one of the transitions between states, as the energy of the transition will be equal to the energy of the photon, whether it is the emitted photon in spontaneous emission, the stimulating photon in stimulated emission or the absorbed photon in the absorption transition.

Example
What is the wavenumber ($${\tilde{\nu}}_{photon}$$) of electromagnetic radiation emitted when a proton in a one-dimensional box with length 472 nm decays from the fifth excited state to the fourth excited state?

SOLUTION: This transition is a spontaneous emission where the energy of the emitted photon can be determined by the difference between the energy levels of the transition,


 * $$\begin{align}

E_{photon} & = \Delta{E} \\ & = E_j - E_i, \\ & \text{where} \ i \ \text{and} \ j \ \text{are the final state and the excited state, respectively.} \\ \end{align}$$

As this transition occurs between the fourth and fifth excited states, $$i$$ and $$j$$ will be 5 and 6, respectively. From the energy levels of a particle in a 1D box,
 * $$\begin{align}

E_6 - E_5 & = \frac{h^2 6^2}{8 m_p L^2} - \frac{h^2 5^2}{8 m_p L^2} \\ & =\frac{11 h^2}{8m_p L^2}, \\ & \text{where} \ m_p \ \text{is the mass of a proton in kg and} \ L \ \text{is the given length of the box in m,} \ (1 \ \text{nm}=1\cdot10^{-9}\text{m})\text{.} \\ \end{align} $$ Substituting these values into the above equation gives,
 * $$\begin{align}

E_{photon} & =\frac{11(6.62607015\cdot10^{-34}\,\text{m}^2\text{kg/s})^2}{8(1.67262192\cdot 10^{-27}\,\text{kg})(472\cdot10^{-9}\,\text{m})^2} \\ & = 1.62 \cdot 10^{-27} \, \text{J} \\ \end{align} $$ Using the wavenumber variation of Planck's relation, the wavenumber of a photon of this exact energy can be determined and expressed in convenient units of $$\text{cm}^{-1}$$.


 * $$\begin{align}

\tilde{\nu}_{photon} & =\frac{E_{photon}}{hc}\text{, where}\ h\ \text{is Planck's constant and}\ c\ \text{is the speed of light expressed in cm/s.} \\ & =\frac{1.62\cdot10^{-27}\,\text{J}}{(6.62607015\cdot10^{-34}\,\text{m}^2\text{kg/s})(2.99792458\cdot10^{10}\,\text{cm/s})} \\ & =8.16 \cdot 10^{-5} \, \text{cm}^{-1} \\ \end{align}$$