Understanding the Boltzmann Distribution - Integrating N(E) from 0 to Infinity

[skp524]

For N(E)=Aexp(-E/kT), I know that N(E) is the no. of particles with a certain energy E,
but why does integrating N(E) from 0 to infinity equal to 1? Although I realize that it means that there is 100% probability to find a particle in this range, I want to know why summing up all no. of particles at all energy levels leading to probability, i.e. 1. Do I have some misunderstanding about this equation?

 

[Andrew Mason]

For N(E)=Aexp(-E/kT), I know that N(E) is the no. of particles with a certain energy E,
but why does integrating N(E) from 0 to infinity equal to 1? Although I realize that it means that there is 100% probability to find a particle in this range, I want to know why summing up all no. of particles at all energy levels leading to probability, i.e. 1. Do I have some misunderstanding about this equation?

You are dealing with a normalized distribution function. The amplitude, A, is chosen such that the area under the graph is equal to 1. Can you work out what that amplitude would be?

AM

 

[chrisbaird]

Technically, N(E) is not the absolute total number of particles at an energy E, but is rather the probability-energy density; in other words, the number of particles as a fraction of the of the total number of particles at a certain energy per unit energy. The quantity N(E)ΔE is the number of particles in energy range ΔE as a fraction of the of the total number of particles. So the sum ΣN(E)ΔE is the number of particles at all energies as a fraction of the total number of particles, which must be one. Let the range ΔE become very small and the sum becomes an integral: ∫N(E)dE = 1.