A source contains two phosphorous radio nuclides `""_15^32P`(T1/2 = 14.3d) and `""_15^33P` (T1/2 = 25.3d). Initially, 10% of the decays come from `""_15^33P`. How long one must wait until 90% do so?
Solution
Half life of `""_15^32P`, T1/2 = 14.3 days
Half life of `""_15^33P`, T’1/2 = 25.3 days
`""_15^33P` nucleus decay is 10% of the total amount of decay.
The source has initially 10% of `""_15^33P` nucleus and 90% of `""_15^33P` nucleus.
Suppose after t days, the source has 10% of `""_15^32P` nucleus and 90% of `""_15^33P` nucleus.
Initially:
Number of `""_15^33P` nucleus = N
Number of `""_15^32P` nucleus = 9 N
Finally:
Number of `""_15^33P nucleus = 9N'`
Number of `""_15^32P nucleus = N'`
For `""_15^32P` nucleus, we can write the number ratio as:
`N'/(9N) = (1/2)^t/T_(1/2)`
`N' =9N (2)^(-1/14.3)`
For `""_15^33P` we can write the number ratio as:
`9N' = N(2)^(-1/(25.3))` ...2
On dividing equation (1) by equation (2), we get:
`1/9 = 9 xx 2^(t/25.3 - t/14.3)`
`1/18 = 2^(-(11t/(25.3 xx 14.3)))`
log 1 - log 81 = -11t/(25.3 xx 14.3) log 2
`(-11t)/(25.3 xx 14.3) = (0 - 1.908)/(0.301)`
t = (25.3 xx 14.3 xx 1.908)/11 xx 0.301 ~~ 208.5 day
Hence, it will take about 208.5 days for 90% decay of `""_15P^33`.
