Question #3594d

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Answer 1 · 2017-04-10T19:12:22

The amount left is $=0.0625g$

The half life of Cobalt 60 is $t_(1/2)=5.25 years$

The radioactive decay constant is $lambda=ln2/(t_(1/2))$

So,

$lambda=ln2/(5.25)=0.69/(5.25 )$

$= 0.1320(years^-1)$

We apply the equation

$A=A_0*e^(-lamdat)$

The activity is proportional to the mass.

$m=m_0*e^(-lamdat)$

$m=1*e^-(0.1314*21)$

$m=1*e^-2.773$

$m=1*0.0625$

$m=0.0625g$

We can do this differently

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Every half life, the amount is divided by $2$

$21/5.25=4$ times

So the initial amount is divided by

$1/16=0.625$