#### Question

A source contains two phosphorous radio nuclides `""_15^32P`(*T*_{1/2 =} 14.3d) and `""_15^33P` (*T*_{1/2} = 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`, *T*_{1/2 =} 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`.