00:01
Hi there! so for this problem we are told that for a certain metal, the stiffness of the interatomic bond and the mass of one atom are such that the spacing of the quantum oscillator energy level, let's call this q, is equal to 1 .5 times 10 to the minus 23 joules, and a nanoparticle of 5 ,000 of 1.
00:32
This metal consists of 10 atoms and has a total internal energy that is also given.
00:39
So the number of atoms, let's call this n .a.
00:46
Is equal to 10th.
00:51
And the total thermal energy, let's call this et, is equal to 18 times 10 to the minus 23 joules.
01:04
So for part a of this problem, we are asked about what is the entropy of this nanoparticle? so in order to calculate the entropy of the nanoparticle, firstly, we should calculate the number of microstates.
01:21
Suppose that all the energy is in the form of the thermal energy.
01:28
And in this case, the number of oscillations then is going to be equal to 30.
01:34
So the number of quanta, number of quanta is equal to the value given for the thermal energy, divided by the energy given for one quantum.
01:55
So that will be 18 times 10 to the minus 23 joules divided by 1 .5 times 10 to the minus 23 joules divided by 1 .5 times 10 to the minus 20.
02:05
23 joules.
02:06
So from this we obtain an approximate value of 12.
02:10
Now the number of microstates that we call, we're going to call just simply a sigma is equal to 12 plus 30 minus 1 and this factorial and this divided by 12 factorial times 30 minus 1 factorial, where 30 is 30 is is the number of oscillators.
02:41
And then from this, we obtain 41 factorial divided by 12 factorial times 29 factorial.
02:53
So if we simplify this expression, we will find that this is equal to 7 .91 times 10 to 9.
03:06
So therefore, the entropy of the nanoparticle is given by the following expression...
