00:01
So first we should find the values for a and b that are asked in the problem and they are corresponding to the probabilities of p of 0 and p of 1e energy state.
00:12
Generally speaking, p of some energy is constant a times exponential function of the minus constant b times the energy.
00:25
So from the figure 10 .2 we can see that for that for p of 0, the probability is actually 0 .385.
00:39
So a is going to be 0 .385 because this is going to be 0 .385 because this is going to be 0.
01:03
The figure and the same figure we should somehow find for the energy 1e now the value of the cost to the b so p is 1a and then the sequence to cost a times b to the minus b and then 1e energy level so when we divide this width a we obtain we obtain it p 1a 1a 1 e over a is actually exponential function to b times 1a and this left hand side has a numerical value and this value is this is from the figure and also all but we already know which is equal to 0 .665 now notice that if we put both sides as as an argument of a natural logarithm function.
02:24
So we do this to both sides of the equation.
02:26
You obtain that basically minus b over 1e is equal to n of 0 .665.
02:44
Therefore, i can say that b equals, since this is going to be negative value, b equals 1e times 0 .41.
02:59
And this will be how we will look and treat this constant in the future.
03:07
So basically what we will say, but we say probability of e is 0 .385 times expression function negative and then 0 .41.
03:25
That energy over 1e as energy.
03:33
And using this and putting into a calculator the proper values, we obtain that.
03:41
For example, for p0 we know the probability.
03:44
For p1e, it would be 0 .385 times e and here we will have negative 0 .411e over 1e, which cancels out and this is equal to 0 .256.
04:06
And in the same manner, we will just have, for example, for 2e, this will be 0 .385, 385 times exponential function.
04:20
Now we will have minus 0 .41 times 2 e's and 1e and this will be just canceling and this will be 2...