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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?

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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.
Updated On: Nov 26, 2025
  • \( e^{-\frac{E_0}{kT}} \)
  • \( 1 - e^{-\frac{E_0}{kT}} \)
  • \( e^{\frac{E_0}{kT}} \)
  • \( 1 + e^{\frac{E_0}{kT}} \)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Boltzmann distribution. In thermodynamics, the Boltzmann distribution quantifies the probability \( P(E) \) that a system's particle possesses energy \( E \): \[P(E) = \frac{e^{-\frac{E}{kT}}}{Z}\] where: - \( E \) represents the particle's energy, - \( k \) is the Boltzmann constant, - \( T \) denotes the system's temperature, - \( Z \) is the partition function, which ensures the probability distribution is normalized. The Boltzmann distribution demonstrates an exponential decrease in the probability of a particle having a given energy as that energy increases. Step 2: Calculating the probability of energy exceeding \( E_0 \). The objective is to determine the probability that a particle's energy surpasses \( E_0 \). This is equivalent to the complement of the probability that the particle's energy is less than or equal to \( E_0 \): \[P(E > E_0) = 1 - P(E \leq E_0)\] Based on the Boltzmann distribution, the probability of a particle having energy less than or equal to \( E_0 \) is computed as: \[P(E \leq E_0) = \int_0^{E_0} \frac{e^{-\frac{E}{kT}}}{Z} dE\] The integral's solution yields the probability of energy being less than \( E_0 \). Consequently, the probability of energy exceeding \( E_0 \) is: \[P(E > E_0) = e^{-\frac{E_0}{kT}}\] Step 3: Conclusion. Therefore, according to the Boltzmann distribution, the probability that a particle will have energy greater than \( E_0 \) is: \[P(E > E_0) = e^{-\frac{E_0}{kT}}\] Answer: The probability is \( e^{-\frac{E_0}{kT}} \).
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