Question

A source contains two phosphorous radio nuclides `""_15^32"P"` (T_{1/2 =} 14.3d) and `""_15^33"P"` (T_1/2 = 25.3d). Initially, 10% of the decays come from `""_15^33"P"`. How long one must wait until 90% do so?

Answer 1

Half life of `""_15^32"P"`, T_{1/2 =} 14.3 days

Half life of `""_15^33"P"`, T’_1/2 = 25.3 days

`""_15^33"P"` nucleus decay is 10% of the total amount of decay.

The source has initially 10% of  `""_15^33"P"` nucleus and 90% of `""_15^33"P"` nucleus.

Suppose after t days, the source has 10% of `""_15^32"P'` nucleus and 90% of `""_15^33"P"` nucleus.

Initially:

Number of `""_15^33"P"` nucleus = N

Number of `""_15^32"P"` nucleus = 9 N

Finally:

Number of `""_15^33"P"` nucleus = 9N

Number of `""_15^32"P"` nucleus = N

For `""_15^32"P"` nucleus, we can write the number ratio as:

`"N'"/(9"N'") = (1/2)^("t"/"T"_(1//2))`

`"N'" = 9"N" (2)^(-1/14.3)`    ....(1)

For `""_15^33"P"` we can write the number ratio as:

`9"N'" = "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^(-((11"t")/(25.3 xx 14.3)))`

log 1 - log 81 = `(-11"t")/(25.3 xx  14.3)` log 2

`(-11"t")/(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 `""_15"P"^33`.