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.

Explanation

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`.