Answer:
Step-by-step explanation:
The decay rate of strontium-90 is -.0244 as given.
For b., we have to use the formula to find out how much is left after 30 years. This will be important for part d.
[tex]A(t)=400e^{-.0244(30)}[/tex] which simplifies a bit to
A(t) = 400(.4809461353) so
A(t) = 192.4 g
For c., we have to find out how long it takes for the initial amount of 400 g to decay to 100:
[tex]100=400e^{-.0244t}[/tex]. Begin by dividing both sides by 400:
[tex].25=e^{-.0244t[/tex] and then take the natural log of both sides:
[tex]ln(.25)=lne^{-.0244t[/tex] . The natural log and the e cancel each other out since they are inverses of one another, leaving us with:
ln(.25) = -.0244t and divide by -.0244:
61.8 years = t
For d., we figured in b that after 30 years, 192.4 g of the element was left, so we can use that to solve for the half-life in a different formula:
[tex]A(t)=A_0(.5)^{\frac{t}{H}[/tex] and we are solving for H. Filling in:
[tex]192.4=400(.5)^{\frac{30}{H}[/tex] and begin by dividing both sides by 400:
[tex].481=(.5)^{\frac{30}{H}[/tex] and take the natural log of both sides, which allows us to pull the exponent out front. I'm going to include that step in with this one:
ln(.481) = [tex]\frac{30}{H}[/tex] ln(.5) and then divide both sides by ln(.5):
[tex]\frac{ln(.481)}{ln(.5)}=\frac{30}{H}[/tex] and cross multiply and isolate the H to get:
[tex]H=\frac{30ln(.5)}{ln(.481)}[/tex] and
H = 28.4 years
