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How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to 5% of its original value?

(At = A0e^-kt)


A 3 days

B 4.6 days

C 5 days

D 10.73 days

E 19.97 days

Verified Answer :

Nirina7

we know that  N=No exp(-kt), where k is the decay constant, 5% of its original value means N=(5/100)No
so, for t=t', N=(5/100)No, (5/100)No=Noexp(-kt'), (5/100)=exp(-kt')
Ln(5/100)=Lnexp(-kt')= - kt',  -2.99 = -0.15t', so t' = 19.93
so the answer is E 19.97 days

Answer:

E. 19.97 days.

Step-by-step explanation:

The function that models this situation is

\(A(t)=A_{0}e^{-kt}\)

This expression models the decay behaviours, where \(k=0.15\) is the decay constant.

Now, 5% of its original value refers to \(A(t)=0.05(A_{0})\).

Using all this information in the formula, we have

\(A(t)=A_{0}e^{-kt}\\0.05(A_{0})=A_{0}e^{-(0.15)t}\\0.05=e^{-(0.15)t}\)

Here, we need to use logarithms to eliminate the power

\(ln(0.05)=ln(e^{-(0.15)t})\\ln(0.05)=-0.15t\\t=\frac{ln(0.05)}{-0.15} \approx 19.97\)

Therefore, the right answer is E. 19.97 days.

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