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#