Co-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of Co–60 atoms will remain in a particular sample after 26.5 years?

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Co-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of Co–60 atoms will remain in a particular sample after 26.5 years?

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Answer 1 · 2017-01-20T16:52:05

The equation for radioactive decay is given by

$A=A_oe^((-0.693t)/T_(1/2))$

where $T_(1/2)$ is the half-life of the material and $t$ is the period of time that elapses in a given case.
In this case, it is actually the ratio $A/(A_o)$ that we are looking for, so this will be found by evaluating the exponential function

$A/(A_o)= e^(((-0.693(26.5))/5.3) = e^(-3.465)=0.03125$

So, the portion remaining is 0.03125 or 3.125 % which is $1/32$ as a fraction.

Note that I used the exponential form of the equation. You could also use the (simpler) form

$A=A_o(1/2)^(t/T_(1/2))$

if you rather.

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Co-60 is a beta emitter with a half-life of 5.3 years. Approximately what fraction of Co–60 atoms will remain in a particular sample after 26.5 years?

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