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A source contains two phosphorus radio n...

A source contains two phosphorus radio nuclides `""_(15)^(32)P (T_(1//2) = 14.3 d) and ""_(15)^(33)P(T_(1//2) = 25.3 d)`. Initially , 10% of the decay come from `""_(15)^(33)P`. How long one must wait until 90% do so?

Text Solution

Verified by Experts

`T_(1//2) "of " ""_(15)^(32)P` is 14.3d
`T_(1//2) " of " ""_(15)^(33)P "is " 25.3d`
`A_(33)(0) = 10% = 1/10 = 0.1`
`A_(33)(0) = 10% = 1/10 = 0.1`
`A_(33) (t) = 90% = 0.4`
10% of activity belong to `""_(15)^(33)P` P out of the total `A_(33)(0) + A_(32)(0)`
i.e., `A_(33)(0) = 0.1 (A_(33)(0) + A_(32) (0)) = 0.1 A_(33)(0) + 0.1A_(32)(0)`
`:. 9 A_(33)(0) =A_(32)(0) " or " (A_(33)(0))/(A_(32)(0)) = 1/9`
At the reqired time t, `A_(33)(t) = 90% = 0.9` (of total)
i.e., `A_(33)(t) = 0.9A_(33)(t) + 0.9A_(32)(t) = 0.9A_(33)(t) + 0.9A_(32)(t)`
`A_(33)(t) = 9A_(32)(t) " " :. (A_(33)(t))/(A_(32)(t)) = 9`
Also , `A_(33)(t) = A_(33)(0)e^(-lambda_33 t)`
`A_(32)(t) = A_(32)(0) e^(-lambda_(32) t)`
`(A_(33)(t))/(A_(32)(t)) = (A_(33)(c ))/(A_(32)(e )) cdot e^(lambda_(32) - lambda_(33))t`
`9 = 1/9 e^(lambda_(32) - lambda_(33)t`
`(lambda_(32) - lambda_(33))t = log_(e ) 81 " " :. t = (log_(e) 81)/((lambda_(32) - lambda_(32)) ........(a)`
`lambda_(32) = (0.6931)/(14.3), lambda_(33) = (0.6931)/(25.3)`
From `(a) t = (log_e 81)/((lambda_(32) - lambda_(33)) ..........(a)`
`lambda_(32) = (0.6931)/(14.3) , lambda_(33) = (0.6931)/(25.3)`
From (a) `t = ((log_e 81)/((0.6931)/(14.3) - (0.6931)/(25.3))) = 208.5 dys`.
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