Question #70676

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Answer 1 · 2017-04-16T13:42:19

The mass left is $=6.39g$

The half life of Cobalt 60 is $t_(1/2)=5.27 years$

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

So,

$lambda=ln2/(5.27)=0.69/(5.27 )$

$= 0.1309(years^-1)$

We apply the equation

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

The activity is proportional to the mass.

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

$m=200*e^-(0.1309*26.3)$

$m=200*e^-3.443$

$m=200*0.03195$

$m=6.391g$