Curium-243 has a half-life of 28.5 days. In a sample of 5.6 grams of curium-243, how many grams will remain after 12 days?

Question

anonymous · Answer

There are two parts to this problem. Here's how to do the first.$$y=Ce^{kt},\text{ where}\
y=\text{amount of substance after time t}\
C=\text{initial amount of substance}\
k=\text{relative decay factor}\
t=\text{time}$$
You're given that Curium-243 has a half-life of 28.5 days, so after this period of time, you have
$$\frac{1}{2}=e^{28.5k}$$And solve for k.

anonymous · Answer

Second part: Using the k you found in the first part, use the next set of information and the formula:
$$y=5.6e^{12k}$$
And solve for y.

anonymous · Answer

i got a really big number i dk what i did wrong

anonymous · Answer

$$\frac{1}{2}=e^{28.5k}\
\ln\left(\frac{1}{2}\right)=28.5k\
k=\frac{\ln\left(\frac{1}{2}\right)}{28.5}\
k\approx-0.0243$$
Did you get this?