Question

The force constant for$\mathbf{\text{HF}}$ is $\mathbf{966}\mathbf{}\mathbf{\text{N}}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{-}\mathbf{1}}$. Using the harmonic oscillator model, calculate the relative population of the first excited state and the ground state at$\mathbf{300}\mathbf{}\mathbf{\text{K}}$ .

Short answer

The relative population of the first excited state to the ground state is $2.4\times {10}^{-9}$.

Step-by-step solution

Given information

The force constant HF and temperature is $k=966{\text{Nm}}^{-1}$ and $T=300$.

Concept of Maxwell–Boltzmann distribution

The Maxwell–Boltzmann distribution is concerned with the energy distribution between identical but distinct particles. It denotes the likelihood of the distribution of states in a system with varying energies. The so-called Maxwell distribution law of molecular velocities is a specific instance.

Calculate the relative probability of molecules

The formulafor the relative probability of finding molecules in the first excited state and in the ground state$n=0$is given by:

$\frac{P\left(1\right)}{P\left(0\right)}=\exp \left(\frac{-[{\epsilon }_n-{\epsilon }_0]}{k_{B}T}\right)$

Or in other words,$\frac{P\left(1\right)}{P\left(0\right)}=\exp \left(\frac{-h\nu }{k_{B}T}\right)$.

Whereis the Boltzmann constant$=1.38\times {10}^{-23}{\text{J}}^{-1}{\text{K}}^{-1}$.

As there are missing some information.

To get the factor to let’s use the formula, $h\nu =\frac{h}{2\pi }\times \sqrt{\frac{k}{\mu }}$.

Calculate the reduced mass

To get thefactor (the reduced mass) use the formula,$\mu =\frac{m_H\times m_F}{m_H+m_F}$.

Where$m_H$and$m_F$are the isotopic masses of the two elements. convert it straight to kilograms from amu so the final formula is shown below.

$\mu =\frac{m_H\times m_F}{m_H+m_F}\times \left(\frac{1\text{g}}{6.022\times {10}^{23}\text{amu}}\right)\times \left(\frac{1\text{kg}}{{10}^3\text{g}}\right)$

Calculate the reduced mass.

$\begin{array}{c}{m}_H=1.007825032\text{amu}\\ m_F=18.9984032\text{amu}\\ \mu =\frac{1.007825032\times 18.9984032}{1.007825032+18.9984032}\times \left(\frac{1\text{g}}{6.022\times {10}^{23}\text{amu}}\right)\times \left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ \mu =1.59\times {10}^{-27}\text{kg}\end{array}$

Calculate the 

From the formula calculateand substitute the values:

$\begin{array}{c}h\nu =\frac{6.625\times {10}^{34}}{2\pi }\times \sqrt{\frac{966}{1.59\times {10}^{-27}}}\\ h\nu =8.22\times {10}^{-20}\text{J}\end{array}$

So, the final calculation is:

$\begin{array}{c}\frac{P\left(1\right)}{P\left(0\right)}=\exp \left(\frac{-8.22\times {10}^{-20}}{1.38\times {10}^{-23}\times 300}\right)\\ \frac{P\left(1\right)}{P\left(0\right)}=2.4\times {10}^{-9}\end{array}$