Update solutions to the lecture 1 excercises.

src/1_einstein_model_solutions.md

@@ -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.