Using Boltzmann distribution law to find Temperature (1% of photons> 1eV)

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SUMMARY

The discussion focuses on using the Boltzmann distribution law to calculate the temperature at which 1.00% of photons exceed an energy of 1.00 eV. The user initially attempted to solve for the temperature at which 1% of photons are at 2 eV, leading to the equation T = -2/(k * log(0.01)). However, the challenge lies in determining the temperature for photons greater than 1 eV, which requires a different approach to the Boltzmann distribution function. The correct formulation involves adjusting the energy levels in the exponential terms to reflect the desired threshold of 1 eV.

PREREQUISITES
  • Understanding of the Boltzmann distribution law
  • Familiarity with photon energy levels and their relation to temperature
  • Basic knowledge of statistical mechanics
  • Proficiency in logarithmic functions and their applications in physics
NEXT STEPS
  • Research the application of the Boltzmann distribution in photon energy calculations
  • Learn how to derive temperature from energy thresholds using statistical mechanics
  • Explore the implications of photon energy distributions in thermodynamic systems
  • Study examples of calculating temperatures for various energy levels using the Boltzmann equation
USEFUL FOR

Students in physics, particularly those studying thermodynamics and statistical mechanics, as well as educators and researchers interested in photon energy distributions and their temperature relationships.

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Homework Statement



Use the Boltzmann distribution function to calculate the temperature at which 1.00% of a population of photons will have energy greater than 1.00 eV. The energy required to excite an atom is on the order of 1 eV.


The Attempt at a Solution



I attached my attempt but it only solves for temperature at which 1% of photons are at 2eV. I have no idea to find temperature at which 1% of photons are greater than 1eV.
 

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\begin{align}1.00\% &= \frac{N_{2eV}}{N}\\1.00\% &= \frac{e^{-2/kT}}{e^{-1/kT} + e^{-2/kT}}\\1.00\% &= e^{-2/kT}\\\log(0.01) &= -2/kT\\T &= -\frac{2}{k\log(0.01)}\end{align}
 

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