Question

Calculate the temperature of a two level system of energy separation equivalent to 400 per centimeter when the population of the upper state is one third of the lower state

Answer 1

The temperature of the two-level system is approximately 0.5235 K.

The Boltzmann distribution describes the population of particles in different energy states at a given temperature. For a two-level system with energy separation
`\( \Delta E \)`, the ratio of the populations
`(\( (N_2)/(N_1) \))` is given by:

`$\[ (N_2)/(N_1) = e^{-(\Delta E)/(kT)} \]`

where:

• 
  `\( N_1 \)` and
  `\( N_2 \)` are the populations of the lower and upper states, respectively.
• 
  `\( \Delta E \)` is the energy separation between the two states.
• 
  `\( k \)` is the Boltzmann constant
  `(\(8.617333262145 * 10^(-5) \, \text{eV/K}\))`.
• 
  `\( T \)` is the temperature in Kelvin.

Given that the energy separation is 400 per centimeter, we need to convert it to electron volts (eV). The conversion factor is
`\(1 \, \text{cm}^(-1) = 1.2398 * 10^(-4) \, \text{eV}\)`.

`\[ \Delta E = 400 \, \text{cm}^(-1) * 1.2398 * 10^(-4) \, \text{eV/cm}^(-1) \]`

`\[ \Delta E = 0.049592 \, \text{eV} \]`

Now, we can set up the equation using the given information that the population of the upper state is one-third of the lower state
`(\( (N_2)/(N_1) = (1)/(3) \))` and solve for T.

`$\[ (1)/(3) = e^{-(\Delta E)/(kT)} \]`

`\[ T = -(\Delta E)/(k \ln\left((1)/(3)\right)) \]`

`$\[ T = -\frac{0.049592 \, \text{eV}}{(8.617333262145 * 10^(-5) \, \text{eV/K}) \ln\left((1)/(3)\right)} \]`

`$\[ T \approx -\frac{0.049592 \, \text{eV}}{(8.617333262145 * 10^(-5) \, \text{eV/K}) * (-1.09861228867)} \]`

`\[ T \approx (0.049592)/(0.09465100485) \]`

`\[ T \approx 0.5235 \, \text{K} \]`

So, the temperature of the two-level system is approximately 0.5235 K.

The complete question: Calculate the temperature of a two-level system of energy separation equivalent to 400 per centimeter when the population of the upper state is one-third of the lower state.