00:01
All right, so for this problem, so for part a, you can actually utilize the result in the problem 43 -34.
00:17
And in there, you have expression for nk, so k is the energy.
00:24
And by utilizing this expression, you can find out the derivative d, nk over dk.
00:34
So the expression gives you 1 .13n over kt, kt 3 over 2, and 1 over 2, k with power 1 over 2, minus k with power 3 over 2, divided by kt, and times exponential negative k, divide by kt, by setting k equal to kp.
01:10
So we need to set this equal to 0.
01:12
Because we want to find the extreme value.
01:16
So by solving this equation, we can find the kp equal half kt.
01:26
And just plugging the value for t equal 1 .5 times 10 to the 7th kelvin.
01:35
So you can see that kp equal 0 .65 kilo.
01:47
Kilo eve, yeah.
01:51
And for part b, so there is a, you can actually use the resulting the problem 1935...