Question:
In a two-level atomic system, the excited state is 0.2 eV above the ground state. Considering the Maxwell-Boltzmann distribution, the temperature at which 2% of the atoms will be in the excited state is _____ K. (up to two decimal places)
(Boltzmann constant \(k_B = 8.62 \times 10^{-5}\) eV/K)
In a two-level atomic system, the excited state is 0.2 eV above the ground state. Considering the Maxwell-Boltzmann distribution, the temperature at which 2% of the atoms will be in the excited state is _____ K. (up to two decimal places)
(Boltzmann constant \(k_B = 8.62 \times 10^{-5}\) eV/K)
(Boltzmann constant \(k_B = 8.62 \times 10^{-5}\) eV/K)
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When solving problems involving Boltzmann statistics, if the percentage of particles in one state is given, you can find the ratio of populations needed for the formula. For a small percentage in the excited state, say p, the ratio \(N_2/N_1\) is approximately \(p/(1-p)\). Be careful with units; ensure that \(\Delta E\) and \(k_B T\) have the same energy units before taking the ratio.
Updated On: Sep 8, 2025
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Correct Answer: 596.17
Solution and Explanation
Step 1: Understanding the Concept:
The population of atomic energy levels at thermal equilibrium is described by the Maxwell-Boltzmann distribution. This distribution relates the ratio of the number of atoms in two different energy states to the energy difference between the states and the temperature of the system.
Step 2: Key Formula or Approach:
The ratio of the number of atoms in an excited state (\(N_2\)) to the number of atoms in the ground state (\(N_1\)) is given by: \[ \frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-\frac{E_2 - E_1}{k_B T}} = \frac{g_2}{g_1} e^{-\frac{\Delta E}{k_B T}} \] where \(g_1\) and \(g_2\) are the degeneracies of the ground and excited states, respectively, \(\Delta E\) is the energy difference, \(k_B\) is the Boltzmann constant, and T is the absolute temperature. Since the degeneracies are not mentioned, we assume they are non-degenerate, so \(g_1=g_2=1\).
Step 3: Detailed Explanation:
We are given that 2% of the atoms are in the excited state. If the total number of atoms is N, then: Number of atoms in the excited state, \(N_2 = 0.02 N\).
Number of atoms in the ground state, \(N_1 = N - N_2 = N - 0.02N = 0.98 N\).
The ratio of the populations is: \[ \frac{N_2}{N_1} = \frac{0.02 N}{0.98 N} = \frac{2}{98} = \frac{1}{49} \] Now, we can use the Boltzmann distribution formula with \(\Delta E = 0.2\) eV and \(g_1=g_2=1\): \[ \frac{1}{49} = e^{-\frac{0.2 \text{ eV}}{k_B T}} \] To solve for T, we take the natural logarithm of both sides: \[ \ln\left(\frac{1}{49}\right) = -\frac{0.2}{k_B T} \] \[ -\ln(49) = -\frac{0.2}{k_B T} \] \[ \ln(49) = \frac{0.2}{k_B T} \] Rearranging the formula to solve for T: \[ T = \frac{0.2}{k_B \ln(49)} \] Now, substitute the given value for \(k_B\): \[ T = \frac{0.2 \text{ eV}}{(8.62 \times 10^{-5} \text{ eV/K}) \times \ln(49)} \] Using \(\ln(49) \approx 3.8918\): \[ T = \frac{0.2}{ (8.62 \times 10^{-5}) \times 3.8918} = \frac{0.2}{0.00033547} \approx 596.17 \text{ K} \] Step 4: Final Answer:
The temperature at which 2% of the atoms will be in the excited state is 596.17 K.
The population of atomic energy levels at thermal equilibrium is described by the Maxwell-Boltzmann distribution. This distribution relates the ratio of the number of atoms in two different energy states to the energy difference between the states and the temperature of the system.
Step 2: Key Formula or Approach:
The ratio of the number of atoms in an excited state (\(N_2\)) to the number of atoms in the ground state (\(N_1\)) is given by: \[ \frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-\frac{E_2 - E_1}{k_B T}} = \frac{g_2}{g_1} e^{-\frac{\Delta E}{k_B T}} \] where \(g_1\) and \(g_2\) are the degeneracies of the ground and excited states, respectively, \(\Delta E\) is the energy difference, \(k_B\) is the Boltzmann constant, and T is the absolute temperature. Since the degeneracies are not mentioned, we assume they are non-degenerate, so \(g_1=g_2=1\).
Step 3: Detailed Explanation:
We are given that 2% of the atoms are in the excited state. If the total number of atoms is N, then: Number of atoms in the excited state, \(N_2 = 0.02 N\).
Number of atoms in the ground state, \(N_1 = N - N_2 = N - 0.02N = 0.98 N\).
The ratio of the populations is: \[ \frac{N_2}{N_1} = \frac{0.02 N}{0.98 N} = \frac{2}{98} = \frac{1}{49} \] Now, we can use the Boltzmann distribution formula with \(\Delta E = 0.2\) eV and \(g_1=g_2=1\): \[ \frac{1}{49} = e^{-\frac{0.2 \text{ eV}}{k_B T}} \] To solve for T, we take the natural logarithm of both sides: \[ \ln\left(\frac{1}{49}\right) = -\frac{0.2}{k_B T} \] \[ -\ln(49) = -\frac{0.2}{k_B T} \] \[ \ln(49) = \frac{0.2}{k_B T} \] Rearranging the formula to solve for T: \[ T = \frac{0.2}{k_B \ln(49)} \] Now, substitute the given value for \(k_B\): \[ T = \frac{0.2 \text{ eV}}{(8.62 \times 10^{-5} \text{ eV/K}) \times \ln(49)} \] Using \(\ln(49) \approx 3.8918\): \[ T = \frac{0.2}{ (8.62 \times 10^{-5}) \times 3.8918} = \frac{0.2}{0.00033547} \approx 596.17 \text{ K} \] Step 4: Final Answer:
The temperature at which 2% of the atoms will be in the excited state is 596.17 K.
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