Question

A microscopic oscillator has its first and second excited states $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{eV}$and $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{eV}$above the ground-state energy. Calculate the Boltzmann factor for the ground state, first excited state, and second excited state, at room temperature.

Short answer

The value of Boltzmann factor for ground state, first excited state, and second excited state, at room temperature are $0$,$0.135$ and $0.67$respectively.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The first excited state of oscillator is,$E_1=0.05\mathrm{eV}$
• The second excited state energy is,$E_2=0.10\mathrm{eV}$

Concept/Significance of harmonic oscillator.

Any system that can be represented by a restorative force and a mass with inertia is referred to as a simple harmonic oscillator.

It shows how the ideas of Kinetic Energy and Potential Energy interact to generate complex, rhythmic behaviour in a box.

Determination of the Boltzmann factor for the ground state, first excited state, and second excited state, at room temperature

According to Boltzmann distribution, the probability to find particle in an energy state is proportional to,

$E\alpha \Omega \left(E\right)\exp \left(\frac{-E}{k_{\mathrm{B}}\mathrm{T}}\right)$

Here, $\mathrm{\Omega }\left(E\right)$is the number of microstates and $\exp \left(\frac{-E}{k_{\mathrm{B}}\mathrm{T}}\right)$is the Boltzmann factor.

The value of $k_{B}T$at room temperature is given by,

$k_{B}T=\frac{1}{40}\mathrm{eV}$

The value of Boltzmann factor for ground state is given by,

$B{F}_0=\exp \left(\frac{-E_0}{k_{\mathrm{B}}\mathrm{T}}\right)$

Here, $E_0$is the ground state energy whose value is 0.

Substitute values in the above,

$$\begin{gathered} B{F}_0=\exp \left(\frac{0}{k_{B}T}\right) \\ =0 \end{gathered}$$

Thus, the Boltzmann factor for ground state is 0.

The value of Boltzmann factor for first excited state is given by,

$B{F}_1=\exp \left(\frac{-E_1}{k_{\mathrm{B}}\mathrm{T}}\right)$

Here,$E_1$is the first excited state energy.

Substitute values in the above,

$$\begin{gathered} B{F}_1=\exp \left(-\frac{0.05\mathrm{eV}}{\left({\displaystyle \frac{1}{40\mathrm{eV}}}\right)}\right) \\ =0.135 \end{gathered}$$

Thus, the Boltzmann factor for first excited state is 0.135.

The Boltzmann factor for second excited state is given by,

$B{F}_{\mathit{2}}=\exp \left(\frac{\mathit{-}{E}_{\mathit{2}}}{{\mathrm{k}}_{B}T}\right)$

Here, $E_{\mathit{2}}$is the second excited state energy.

Substitute values in the above,

$$\begin{gathered} B{F}_{\mathit{2}}\mathit{=}\exp \left(-\frac{0.10\mathrm{eV}}{\left({\displaystyle \frac{1}{40}\mathrm{eV}}\right)}\right) \\ =0.67 \end{gathered}$$

Thus, the Boltzmann factor for first excited state is 0.67.