Question

A system consists of 110 particlesEach particle can have only two non-degenerate energy states $\epsilon_1$ and $\epsilon_2$ with $\epsilon_2 > \epsilon_1$The energy difference between these two states is 0.10 eVThe temperature at which only 10 particles of the system will be present in the energy state $\epsilon_2$ is _ _ _ K. (rounded off to two decimal places)

Hint:

For non-degenerate two-level systems, use $\frac{N_2}{N_1}=e^{-\Delta E/kT}$ directly

Answer 1

Correct Answer: 503.86

Step 1: Use Boltzmann distribution.
For two non-degenerate energy states,
\[ \frac{N_2}{N_1} = e^{-\frac{\Delta E}{kT}} \]

Step 2: Identify number of particles in each state.
Total particles = 110
Particles in higher energy state $\epsilon_2$ = 10
Therefore, particles in lower energy state $\epsilon_1$ = 100
\[ \frac{N_2}{N_1} = \frac{10}{100} = 0.1 \]

Step 3: Substitute in Boltzmann equation.
\[ 0.1 = e^{-\frac{\Delta E}{kT}} \]
Taking natural logarithm,
\[ \ln(0.1) = -\frac{\Delta E}{kT} \]
\[ \ln(10) = \frac{\Delta E}{kT} \]

Step 4: Convert energy into joule.
\[ \Delta E = 0.10 \text{ eV} \]
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
\[ \Delta E = 0.10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-20} \text{ J} \]

Step 5: Calculate temperature.
\[ T = \frac{\Delta E}{k\ln 10} \]
\[ T = \frac{1.6 \times 10^{-20}}{(1.38 \times 10^{-23})(2.303)} \]
\[ T = 503.86 \text{ K} \]

Step 6: Conclusion.
\[ \boxed{503.86 \text{ K}} \]