Question #3663

I am trying to figure out how to do this question. The half-life of a substance is 120 days. The initial quantity is 72g.
A) Write the specific equation in the form A=A[sub]o[/sub] e[sup]kt[/sup].
B) How much will be left after 85 days?
C) How long until only 8 g is left?

Expert's answer

a. 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=A0e^{- 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

240ln(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=A0e

b. A= 72e

c. 8= 72e

1/9=72e

-2ln(3)= - ln(2)t/120

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

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