Skip to content
Snippets Groups Projects
Commit a9e01e73 authored by Radoica Draškić's avatar Radoica Draškić
Browse files

Merge branch 'radras-master-patch-22627' into 'master'

Update solutions to the lecture 1 excercises.

See merge request !61
parents eda3e272 773806a6
No related branches found
No related tags found
1 merge request!61Update solutions to the lecture 1 excercises.
Pipeline #28061 passed
......@@ -74,12 +74,26 @@ Use the formula $\omega = \sqrt{\frac{k}{m}}$.
2.
$E = N_{^6Li}\hbar\omega_{^6Li}(2 + 1/2)+N_{^7Li}\hbar\omega_{^7Li}(4 + 1/2)$.
Energy per atom is given by
$$
E = \frac{N_{^6Li}}{N}\hbar\omega_{^6Li}(2 + 1/2) + \frac{N_{^7Li}}{N}\hbar\omega_{^7Li}(4 + 1/2).
$$
3.
$E = N_{^6Li}\hbar\omega_{^6Li}\left(n_B(\beta\hbar\omega_{^6Li}) + \frac{1}{2}\right) + N_{^7Li}\hbar\omega_{^7Li}\left(n_B(\beta\hbar\omega_{^7Li}) + \frac{1}{2}\right)$.
Energy per atom is given by
$$
E = \frac{N_{^6Li}}{N}\hbar\omega_{^6Li}\left(n_B(\beta\hbar\omega_{^6Li}) + \frac{1}{2}\right) + \frac{N_{^7Li}}{N}\hbar\omega_{^7Li}\left(n_B(\beta\hbar\omega_{^7Li}) + \frac{1}{2}\right).
$$
4.
$C = \frac{N_{^6Li}}{N}C_{^6Li} + \frac{N_{^7Li}}{N}C_{^7Li}$ where the heat capacities are calculated with the formula from Excercise 2.4.
Heat capacity per atom is given by
$$
C = \frac{N_{^6Li}}{N}C_{^6Li} + \frac{N_{^7Li}}{N}C_{^7Li},
$$
where the heat capacities are calculated with the formula from Excercise 2.4.
0% Loading or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment