# Boltzmann distribution

1. Jul 30, 2011

### 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?

2. Jul 30, 2011

### Andrew Mason

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

3. Aug 1, 2011

### 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.