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.

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,$E_2=0.10\mathrm{eV}$

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.

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.