Question

Thorium -228 is a radioactive substance that decays 50% every 1.9 years. How much of a 50 gram sample will exist after 22.8 years? Round to 3 decimal places. Use the formula A=A0e^ -kt.
A.) 19.34 grams
B.) 30.32 grams
C.) 0.124 grams
D.) 12.4 grams
E.) 7.48 grams

Answer 1

Answer:

$A=0.0122\ gr$

Step-by-step explanation:

The exponential decrease formula that we must use is:

$A=A_0e^ {-kt}$

Where A is the amount of thorium as a function of time

$A_0$ is the initial amount and k is the decrease rate

We know that initially there are 50 gr and that you decay 50% every 1.9 years.

So

$A=A_0*0.5=50*0.5 = 25\\\\t=1.9$

$25=50e^ {-k(1.9)}$

We solve the equation for k

$0.5=e^ {-k(1.9)}$

$ln(0.5)=ln(e^ {-k(1.9)})$

$ln(0.5)=-k(1.9)$

$-1.9k=ln(0.5)$

$k=-\frac{ln(0.5)}{1.9}\\\\k=0.3648$

After 22.8 years the amount of thorium that exists is:

$A=50e^ {-0.3648(22.8)}$

$A=0.0122\ gr$