Over all the possible states in equilibrium, More than one distinct quantum stat
ID: 2304380 • Letter: O
Question
Over all the possible states in equilibrium,
More than one distinct quantum state of a system can have a particular value of energy. Here we remember to count all the states in figuring out the probability that a small system has a particular energy value.
The molecule can be in one of six states. It exchanges energy with a thermal reservoir of other molecules at a temperature T. State 1 has EA, states 2 and 3 each have energy EB = 3EA, and states 4,5, and 6 each have energy EC =5EA. Now to simplify the calculations, let's set the temperature so that kBT=EA.
What is the probability that the molecule has energy EB?
P(EB) =
what is the most probable energy of the first oscillator?
Explanation / Answer
From the given data
total number of given states = 6
Ea = 1 state
Eb = 3Ea = 2 states
Ec = 5Ea = 3 states
hence
probability that the energy of the system is Eb is
P(Eb) = 2/(1 + 2 + 3) = 2/6 = 1/3 = 0.333
also, most probable energy of the osscilator is E
E = sum of( Ea*n) / sum of (n) [ where n is number of states]
E = (Ea + 2*3Ea + 3*5Ea)/(1 + 2 + 3) = 17Ea/6
now, Ea = kbT
hence
E = 17kbT/6 is th emost probable energy of the first osscilator
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