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Question
A source contains two phosphorous radio nuclides `""_15^32"P"` (T1/2 = 14.3d) and `""_15^33"P"` (T1/2 = 25.3d). Initially, 10% of the decays come from `""_15^33"P"`. How long one must wait until 90% do so?
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Solution
Half life of `""_15^32"P"`, T1/2 = 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`.
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