Probability / Energy question

[raintrek]

Homework Statement

- Initial conformational potential energy = 2500 J
- NVT Metropolis Monte Carlo simulation, T = 300 K
- New conformational potential energy = 5000 J

What's the probability this will be accepted within the Metropolis scheme?

Homework Equations

Acceptance Probability:
$$P=exp\left(\frac{-\Delta U_{mn}}{kT}\right)$$

$$\Delta U_{mn}=U(q_{n}^{N})-U(q_{m}^{N})$$

where $$U(q_{m}^{N})$$ is our initial potential and $$U(q_{n}^{N})$$ is our final potential.

The Attempt at a Solution

I'm plugging these numbers in, but the exponent is going to zero, which I know to be wrong. Is there some conversion of energy I'm missing? Seems a bit odd...

 

[Jhero]

Hello, thank you for your post. Based on the given information, the probability of acceptance within the Metropolis scheme can be calculated as follows:

P = exp(-ΔU/kT)

Where:
ΔU = U(qnN) - U(qmN)
k = Boltzmann constant (1.38 x 10^-23 J/K)
T = temperature (300 K)

Substituting the given values, we get:

P = exp(-(5000 J - 2500 J)/(1.38 x 10^-23 J/K * 300 K))
= exp(-2500 J/(4.14 x 10^-21 J))
= exp(-6.04 x 10^-18)

Therefore, the probability of acceptance is:

P = 1 - exp(-6.04 x 10^-18)
= 1 - 0.999999999999999994

Thus, the probability of acceptance is approximately 1, meaning that the new conformational potential energy is highly likely to be accepted within the Metropolis scheme. I hope this helps!