Question #3676

The half-life of a substance is 120 days. The initial quantity is 72g.
a. Write the specific equation in the form A=Aoekt
b. How much will be left after 85 days?
c. How long until only 8 g is left?

Expert's answer

A=A_{0}e^{-kt}.

1/2 A_{0}= A_{0}e^{-kτ}

Where τ is a half-life period. So we get

K=ln(2)/ τ

We have

A=A_{0}e^{- ln(2)t/ τ} = 72e^{- ln(2)t/120}.

b.

A= 72e^{- ln(2)85/120} = 44.17 g.

c.

8= 72e^{- ln(2)t/120} -> 1/9=72e^{- ln(2)t/120} -> -2ln(3)= - ln(2)t/120

240 ln(3)/ln(2) = 379 days.

1/2 A

Where τ is a half-life period. So we get

K=ln(2)/ τ

We have

A=A

b.

A= 72e

c.

8= 72e

240 ln(3)/ln(2) = 379 days.

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