1. The problem statement, all variables and given/known data
If you have 500g of Thorium-234 at t=0. Calculate the initial activity and the activity after 2 weeks. The half life of thorium-234 is 24 days.

2. Relevant equations
$A=A_0e^{-\lambda t} \\ A_0=\lambda N_0 \\ \lambda=\frac{ln2}{t_{\frac{1}{2}}}$

3. The attempt at a solution
For the intial activity I first found lambda by converting 24 days into seconds.

$\lambda=\frac{ln2}{60 \times 60 \times 24 \times 24}=\frac{ln2}{2073600}=3.34 \times 10^{-7} \\$

Then used the second relevant equation listed above by calculating N_0 with the known data (converting to number of moles and then times by Avogadros number).
$A_0=\lambda N_0=(3.34 \times 10^{-7})(\frac{500}{234})(6.02 \times 10^{23})=4.29 \times 10^{17} Bq$

Then after knowing A_0 i can calculate the activity after 2 weeks, with the first relevant equation listed but converting 2 weeks into seconds.
$A=A_0e^{-\lambda t}=(4.29 \times 10^{17})e^{(3.34 \times 10^{-7})(1209600)}=2.86 \times 10^{17} Bq$

My problem is with my answers, they seem way way too high. I was told in a lecture it is very unusual to ever go above a micro Bq.

http://ift.tt/1h7UsOC