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Radioisotopes of phosphorus p^(32) and p^(35)are mixed in the ratio of 2:1 of atoms. The activity of the sample is 2 Ci. Find the activity of the sample after 30 days. T_(1//2)" of "P^(32) is 14 days and T_(1//2)" of "P^(35) is 25 days. |
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Answer» Solution :Let `A_(0)=` initial activity of sample. `A_(10)=` initial activity of isotope 1 and `A_(20)=` initial activity of isotope 2. `A_(0)=A_(10)=A_(20)` Similarly for final activity (Activity after time t) `A_(t)=A_(1t)=A_(2t)` `rArr""A_(t)=A_(10)e^(-lambda_(1)t)+A_(20)e^(-lambda_(2)t)` Now, in the givne equation, `A_(0)=2Ci rArr A_(0)=A_(10)+A_(20)=2"...(i)"` Initial RATIO of ATOMS of isotopes `=2:1` From definition of activity, `A=lambdaN` `rArr""(A_(10))/(A_(20))=(lambda_(1)N_(10))/(lambda_(2)N_(20))=(N_(10))/(N_(20))xx(T_(2))/(T_(1))` where T represents half life `rArr (A_(10))/(A_(20))=(2)/(1)xx(25)/(14)=(50)/(14)=(25)/(7)"...(ii)"` `A_(20)=(7)/(16) and A_(10)=(25)/(16)` `A_(t)=A_(10)e^(-lambda_(1)t)+A_(20)e^(-lambda_(2)t)` `rArr A_(t)=(25)/(16)e^(-(0.693)/(14)xx30)+(7)/(16)e^(-(0.693)/(25)xx30)` Consider the first exponential TERM : `e^(-(0.693xx30)/(14))=e^(-1.485)` Let `y=e^(-1.485)rArr ln y=-1.485` `rArr""logy=(-1.485)/(2.303)rArr y="antilog"((-1.485)/(2.303))` So, from above calculations you can derive a general result i.e., `e^(-X)=" antilog "((-x)/(2.303))` `A_(t)=(25)/(16)xx0.2265+(7)/(16)xx0.4365+(7)/(16)x0.4354=0.5444Ci`. |
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