- #1
teddd
- 62
- 0
Hi everyone!
Here's my problem of the day:
Let's take a box containing 3 identical (but distinguishable) particles A B C. Let this be a canonical ensamble.
Suppose that A has energy \varepsilon_0 and both B and C have energy \varepsilon_1. We thereforre have 2 energy level, n_0,n_1. Take the number of states g_{\alpha} in each energy level \varepsilon_{\alpha} to be 1.
Now, I want to calculate in how many ways the set of population \vec{n}=(n_0,n_1) can be realized.
At first sight I'd say that they're two: I can take (A,BC) or (A,CB), being the particle distinguishable.
But if I use the well-known Boltzmann forumula W(\vec{n})=N!\prod_{\alpha}\frac{ g_{\alpha}^{n_{\alpha}}}{n_{\alpha}} and I put in the g's and n's I've taken above I get:W(\vec{n})=3! \left(\frac{1^1}{1!}\frac{1^2}{2!}\right)=3so there should be three ways to set up the vector \vec{n}!
Where am I mistaking?? Thanks for help!
Here's my problem of the day:
Let's take a box containing 3 identical (but distinguishable) particles A B C. Let this be a canonical ensamble.
Suppose that A has energy \varepsilon_0 and both B and C have energy \varepsilon_1. We thereforre have 2 energy level, n_0,n_1. Take the number of states g_{\alpha} in each energy level \varepsilon_{\alpha} to be 1.
Now, I want to calculate in how many ways the set of population \vec{n}=(n_0,n_1) can be realized.
At first sight I'd say that they're two: I can take (A,BC) or (A,CB), being the particle distinguishable.
But if I use the well-known Boltzmann forumula W(\vec{n})=N!\prod_{\alpha}\frac{ g_{\alpha}^{n_{\alpha}}}{n_{\alpha}} and I put in the g's and n's I've taken above I get:W(\vec{n})=3! \left(\frac{1^1}{1!}\frac{1^2}{2!}\right)=3so there should be three ways to set up the vector \vec{n}!
Where am I mistaking?? Thanks for help!
