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? - Physics

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Numerical

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% ?

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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 Nremains 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 Nremains 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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Chapter 13: Nuclei - Exercise [Page 462]

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