Here's the solution to HW8.
There's an error on the homework solution though. In 1b and 1c, the solution uses 3.3 * 10^4 as activation energy, when in fact it should use 2.3 * 10^4. So, the correct answer for 1b, following Professor Morris' approximation of 1eV ~ 10^4 K, is D ~ 7.1*10^-9 cm^2/s, and t ~ 0.7025 s.
Similarly, for 1c, the correct answer is D ~ 4*10^-20 cm^2/s, and
Alternatively, you could also choose not to approximate 1eV ~ 10^4 K and actually convert the eV unit accurately. But the point is, I don't see any sense in approximating 2.3eV as 3.3* 10^4 K for the activation energy of 1b and 1c.
Anyway, the point distribution breakdown goes as:
1 a) b) c) 10 pts each.
2 a) 5 pts b) 5 pts c) 10 pts
3 a) 10 pts b) 10 pts c) 5 pts
4 a) 5 pts b) 5 pts c) 5 pts
5 a) 5 pts b) 5 pts