$Q(t) = Q_(o)e^(-kt)$ may be used to model radioactive decay. $Q$ represents the quantity remaining after years; $k$ is the decay constant, 0.00011. How long, in years, will it take for a quantity of plutonium-240 to decay to 25% of its original amount?

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$Q(t) = Q_(o)e^(-kt)$ may be used to model radioactive decay. $Q$ represents the quantity remaining after years; $k$ is the decay constant, 0.00011. How long, in years, will it take for a quantity of plutonium-240 to decay to 25% of its original amount?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

The initial quantity at t = 0 is:

$Q_0(0)=Q_0e^(-0.00011*0)=Q_0$

The amount left after t years is one quarter of the initial:

$Q(t)=Q_0/4$

Substitute this expression into the function and the $Q_0$'s will cancel out:

$Q(t)=Q_0/4=Q_0e^(-0.00011t)$

$rArrcancel(Q_0)/(4cancel(Q_0))=e^(-0.00011t)$

$1/4=e^(-0.00011t)$

Take the natural log of both sides, this is the inverse operation of $e^x$ and will undo it (they cancel out):

$ln(1/4)=cancel(ln)cancel(e)^(-0.00011t)$

$ln(1/4)=-0.00011t$

$ln(1/4)=ln(4^-1)=-1*ln(4)$

$t=ln(4)/0.00011~~12600$ years

According to , the half-life of plutonium-240 is 6560 years.
The amount left after $n$ half-lives is $1/n^2$. So after $n=2$ half-lives we will have $1/4$ of the initial amount, which would be 13120 years. Our answer is in the ball park.

$Q(t) = Q_(o)e^(-kt)$ may be used to model radioactive decay. $Q$ represents the quantity remaining after years; $k$ is the decay constant, 0.00011. How long, in years, will it take for a quantity of plutonium-240 to decay to 25% of its original amount?

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