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Questions
A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?
A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3.125% ?
Solution 1
Half-life of the radioactive isotope = T years
Original amount of the radioactive isotope = N0
a) After decay, the amount of the radioactive isotope = N
It is given that only 3.125% of N0 remains after decay. Hence, we can write
`"N"/"N"_0` = 3.125% = `3.125/100 = 1/32`
But `"N"/"N"_0 = "e"^(-lambda"t")`
Where,
λ = Decay constant
t = Time
∴ `-lambda"t" = 1/32`
`-lambda"t" =ln 1 - ln32`
`-lambda"t" = 0 - 3.4657`
`"t" = 3.4657/lambda`
Since `lambda = 0.693/"T"`
∴ `"t" = 3.466/(0.693/"T") ≈ 5"T years"`
Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.
b) After decay, the amount of the radioactive isotope = N
It is given that only 1% of N0 remains after decay. Hence, we can write:
`"N"/"N"_0` = 1% = `1/100`
But `"N"/"N"_0 = "e"^(-lambda"t")`
∴ `"e"^(-lambda"t") = 1/100`
`-lambda"t" = ln 1 - ln 100`
`"t" = 4.6052/lambda`
Since `lambda = 0.693/"T"`
∴ `t = 4.6052/(0.693/"T") = 6.645 "T years"`
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.
Solution 2
`A_0 = 100`
`A_1 = 3.125`
`A_t = A_0.e^(-lambdat)`
`3.125 = 100e^(-0.693/10) t`
`3.125/100 = e^(-0.0693 t)`
`100/3.125 = e^(0.0693t)`
`log_e 100/3.125 = 0.0693t`
`t_2 = (2.303log_10^32)/0.0693`
t = 50.1 year
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