more math problems half life?

Question

The half-life if Actinium (227 Ac) is 22 years. What percent of a present amount of radioactive actinium will remain after 19 years?

Answer 1

One halflife equation used is
P= Po (0.5^ (t/k)), where
P= amount remaining
Po= initial amount
t= time
k= half-life

So for your question, Po= 100%, t=19, k=22
P= 100 (0.5^ (19/22))
P= 54.9566% or ~55% remaining after 19 years (this makes sense because the half-life is 22 years, so the percentage remaining should be > 50%)

Answer 2

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RE:
more math problems half life?
The half-life if Actinium (227 Ac) is 22 years. What percent of a present amount of radioactive actinium will remain after 19 years?

Answer 3

Hi,

y = 100(.5)^(x/22)
y = 100(.5)^(19/22)
y = 54.96% will still remain.

I hope that helps!! :-)

Answer 4

The basic equation is a=e^(-kt), giving the amount left after t years.

First, we must determine k. To do this, we know .5=e^(-22k). Taking the natural log of both sides givess ln(.5) = -22k. Solving for k gives .0315.

Now, plug in 19 for t, giving e^(-19(.0315)) = .5496. So, the percent left is 54.96.

Answer 5

y = 100(.5)^(x/22)
y = 100(.5)^(19/22)
y = 54.96%

Answer 6

The percentage remaining after x years is 0.5^(x/22). After 19 years, approximately 55% remains.
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