Probability of particle overcoming an energy barrier

Question

I'm reading this article called:An experiment to demonstrate the canonical distribution(by M. D. Sturge and Song Bac Toha) Department of Physics, Dartmouth College, Hanover, New Hampshire 03755.

They talked about the probability of a particle overcoming and energy barrier of height $\Delta E$, they say is proportional to $\int_{0}^{\infty}g(\epsilon)e^{-(\epsilon+\Delta E)/kT}d\epsilon$

I have 2 questions:

-Where does this come from, why would integrating that result in a probability, wouldn't I obtain the total number of particles.

-what factor must I include in order for this to be an equality , that is$ P(\Delta E)= factor \int_{0}^{\infty}g(\epsilon)e^{-(\epsilon+\Delta E)/kT}d\epsilon$

Answer 1

According to Boltzman distribution, the probability of finding the particle in a energy $E$ is propotion to $exp(-\frac{E}{KT})$. Consider a degeneracy for energy $E$ is $g(E)$ the density of states, which means that there are $g(E) dE$ independent levels have their energy in between $E$ and $E+dE$, the resultant probability of finding a particle in one of these energy levels is:

$$ p(E)dE \propto g(E) dE e^{-\frac{E}{KT}} $$

Now, there is a energy valley with a barrier $\Delta E$. For particle being able to cross the barrier, the particle must have energy greater that $\Delta E$. The total probability of such particle ($E > \Delta E$):

$$ p(>\Delta E) \propto \int_{\Delta E}^\infty g(\epsilon) d\epsilon e^{-\frac{\epsilon}{KT}} =\int_{0}^\infty g(\epsilon+\Delta E) d\epsilon e^{-\frac{\epsilon + \Delta E}{KT}} $$

(In your expression, the variable inside the density of states $g$ is not correct.)

To make it an equal sign:

$$ p(>\Delta E) = \frac{1}{N} \int_{0}^\infty g(\epsilon+\Delta E) d\epsilon e^{-\frac{\epsilon + \Delta E}{KT}} $$

The normalization constant:

$$ N = \int_0^\infty g(\epsilon) d\epsilon e^{-\frac{\epsilon}{KT}} $$