I did it based on the equation (1+x)^-1=1-x. Therefore, (1+e^-BdeltaE)^-1 is equal to 1-e^-BdeltaE. The denominator is (1+e^-BdeltaE) and not (1-e^-BdeltaE) . I made a mistake, sorry.
A system has two non-degenerate energy levels E1 and E2, where E2>E1>0. The system is at tempreture T. The Average energy of the system is = E1+E2e^(-B*deltaE) / 1+e^(-B*deltaE) where deltaE= E2 -E1 and B=1/kT (k=Boltzmann constant). show that for very low temperatures kT<<deltaE, average...
yes, for the first question(the probability that the particle is in the ground state) I get 0.032. e^(-3.1/500k) / e^(-3.1/500k) + 3e^(-3.0/500k). The answer should be 0.7725
A certain particle is interacting with a reservoir at 500 k and can be in any four possible states. The ground state has energy 3.1 eV and three excited states all have the same energy. what is the probability that the particle is in ground state? what is the probability that the particle is in...
we did it using du nouy ring method. we calibrated the instrument with water and then measured the surface tension of each solution in an increasing order. the solutions were prepared from before.
we did it using du nouy ring method. we calibrated the instrument with water and then measured the surface tension of each solution in an increasing order. the solutions were prepared from before.
We have n-butanol solutions ranging from 0.1 to 0.8 M, which we are supposed to measure their surface tensions. We have to start from the most dilute solution and go up in concentration until we reach 0.8 M. We cannot start from the most concentrated one and go to the most dilute one or just...
when we try to measure surface tension of different concentrated solutions of n-butanol with a tensiometer, we have to go from lower to higher concentration and not the other way around. what is the reason for that?