Answer :

At time t=0, A˳= 8.0 × 108 dis/s

(a)











(b) For large values of time, the value of λ will be the slope negative of the slope of the curve.


λ = 0.028s-1


So, t1/2= 24.4 s


(c)


(d)



(e) The half-life of 108Ag from the graph is 144s.


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