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How much of a radioactive kind of americium will be left after 48 hours if the half-life is 16 hours and you start with 82,112 grams?

User TwoLeftFeet
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1 Answer

17 votes
17 votes

ANSWER

10264 grams

Step-by-step explanation

We want to find how much of americium will be left after 48 hours if its half life is 16 hours and it starts with 82,112 grams.

To do this, we apply the formula for radioactive decay and half life:


\begin{gathered} A=A_o((1)/(2))^{(t)/(T)} \\ \text{where t = time elapsed} \\ T=\text{half life} \\ A_o=\text{initial amount} \end{gathered}

We are given that:

Ao = 82,112 grams

t = 48 hours

T = 16

Therefore, we have that the amount left is:


\begin{gathered} A=82112\cdot((1)/(2))^{(48)/(16)}=82112\cdot((1)/(2))^3 \\ A=10264\text{ grams} \end{gathered}

That is the amount that will be left after 48 hours.

User Rycochet
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