Question

For a system of particles in thermal equilibrium, what is the probability that a particle will have energy greater than \( E_0 \) according to the Boltzmann distribution?

• A. \( e^{-\frac{E_0}{kT}} \)
• B. \( 1 - e^{-\frac{E_0}{kT}} \)
• C. \( e^{\frac{E_0}{kT}} \)
• D. \( 1 + e^{\frac{E_0}{kT}} \)

Hint:

The Boltzmann distribution describes the probability of a system's particle having a particular energy. The probability decreases exponentially with energy, and the partition function \( Z \) is used to normalize the distribution.

Answer 1

The Correct Option is A

Step 1: Understanding the Boltzmann distribution. In thermodynamics, the probability \( P(E) \) that a system's particle has energy \( E \) is given by the Boltzmann distribution: \[ P(E) = \frac{e^{-\frac{E}{kT}}}{Z} \] where: - \( E \) is the energy of the particle, - \( k \) is the Boltzmann constant, - \( T \) is the temperature of the system, - \( Z \) is the partition function, which normalizes the probability distribution. The Boltzmann distribution shows how the probability of a particle having a certain energy decreases exponentially with increasing energy. Step 2: Finding the probability of energy greater than \( E_0 \). We are interested in the probability that the energy of a particle is greater than \( E_0 \). This probability is the complement of the probability that the particle has energy less than or equal to \( E_0 \): \[ P(E > E_0) = 1 - P(E \leq E_0) \] From the Boltzmann distribution, the probability that the particle has energy less than or equal to \( E_0 \) is: \[ P(E \leq E_0) = \int_0^{E_0} \frac{e^{-\frac{E}{kT}}}{Z} dE \] The result of this integral gives the probability that the energy is less than \( E_0 \), and hence the probability that the energy is greater than \( E_0 \) is: \[ P(E > E_0) = e^{-\frac{E_0}{kT}} \] Step 3: Conclusion. Thus, the probability that a particle will have energy greater than \( E_0 \) according to the Boltzmann distribution is: \[ P(E > E_0) = e^{-\frac{E_0}{kT}} \] Answer: Therefore, the probability is \( e^{-\frac{E_0}{kT}} \).