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Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)?
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Hydrogen atoms in the atmosphere of a star are in thermal equilibrium,...
Understanding Thermal Equilibrium in Stars
In a star's atmosphere, hydrogen atoms achieve thermal equilibrium, meaning their energy distribution corresponds to the temperature of the environment. Given an average kinetic energy of 1 eV, we can analyze the population of energy states.
Energy Levels of Hydrogen Atoms
- The energy levels of hydrogen are quantized, given by En = -13.6 eV/n², where n is the principal quantum number.
- For the ground state (n=1), E1 = -13.6 eV.
- For the first excited state (n=2), E2 = -3.4 eV.
- For the second excited state (n=3), E3 = -1.51 eV.
Calculating the Population Ratio
Using the Boltzmann distribution, the population of states can be expressed as:
- The ratio of populations for states n and m is given by: P(n)/P(m) = (g(n)/g(m)) * e^(-(E(n)-E(m))/kT), where g(n) is the degeneracy of the state.
- For hydrogen, the degeneracy g(n) is n². Thus:
- g(1) = 1² = 1
- g(3) = 3² = 9
Applying the Boltzmann Factor
- The energy difference between ground state (n=1) and second excited state (n=3) is E(3) - E(1) = 12.09 eV.
- The temperature T can be derived from the average kinetic energy (1 eV).
- The ratio becomes:
P(3)/P(1) = (g(3)/g(1)) * e^(-12.09 eV/kT)
Conclusion
Given the high energy difference, P(3)/P(1) will be very small, indicating that very few hydrogen atoms are in the second excited state compared to the ground state. Thus, the population in the excited state is significantly lower due to the thermal distribution of energy.
In a star's atmosphere, hydrogen atoms achieve thermal equilibrium, meaning their energy distribution corresponds to the temperature of the environment. Given an average kinetic energy of 1 eV, we can analyze the population of energy states.
Energy Levels of Hydrogen Atoms
- The energy levels of hydrogen are quantized, given by En = -13.6 eV/n², where n is the principal quantum number.
- For the ground state (n=1), E1 = -13.6 eV.
- For the first excited state (n=2), E2 = -3.4 eV.
- For the second excited state (n=3), E3 = -1.51 eV.
Calculating the Population Ratio
Using the Boltzmann distribution, the population of states can be expressed as:
- The ratio of populations for states n and m is given by: P(n)/P(m) = (g(n)/g(m)) * e^(-(E(n)-E(m))/kT), where g(n) is the degeneracy of the state.
- For hydrogen, the degeneracy g(n) is n². Thus:
- g(1) = 1² = 1
- g(3) = 3² = 9
Applying the Boltzmann Factor
- The energy difference between ground state (n=1) and second excited state (n=3) is E(3) - E(1) = 12.09 eV.
- The temperature T can be derived from the average kinetic energy (1 eV).
- The ratio becomes:
P(3)/P(1) = (g(3)/g(1)) * e^(-12.09 eV/kT)
Conclusion
Given the high energy difference, P(3)/P(1) will be very small, indicating that very few hydrogen atoms are in the second excited state compared to the ground state. Thus, the population in the excited state is significantly lower due to the thermal distribution of energy.
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Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)?
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Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)?.
Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hydrogen atoms in the atmosphere of a star are in thermal equilibrium, with an average kinetic energy of 1 eV. The ratio of the number of hydrogen atoms in the 2nd excited state (3) to the number in the ground state (n-1)?.
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