Question

Calculate the relative populations of two quantum states separated by an energy of $0.4\times {10}^{-21}\text{J}$ if the temperature is$25°C$ .

Short answer

The relative population of two quantum states are$0.91$.

Step-by-step solution

Given information

The energy difference between two quantum states and temperature is given as:

hv $=0.4\times {10}^{-21}ȷ$

T$={25}^{°}$

$\text{C}=298\text{K}$

Concept of Maxwell Boltzmann’s constant

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

The relative probability of finding molecules in the excited state $\eta$ and in the ground state $\eta =\circ$ is given as shown below.

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

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

Where $k_B$ is the Boltzmann constant $=1.38\times {10}^{-23}{\text{J}}^{-1}{\text{K}}^{-1}$

Further simplify the equation.

$$\begin{gathered} \frac{P\left(1\right)}{P\left(0\right)}=\exp \left(\frac{-0.4\times {10}^{-21}}{1.38\times {10}^{-23}\times 298}\right) \\ \frac{P\left(1\right)}{P\left(0\right)}=0.91 \end{gathered}$$

So, conclude that the relative population of two energy levels are 0.91

And the ratio is,$\frac{N_{\text{high}}}{N_{\text{low}}}=0.91$