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#