Question

Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample, 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?

Answer 1

Let N be the sample radioactive nuclei at any time t and N₀ be the radioactive nuclei at the initial time.

Given `"dN"/"dt"` = – kN

Where k > 0 is a constant

The equation can be written as

`"dN"/"N"` = – kdt

Taking integration on both sides, we get

`int "dN"/"N" = int - "kdt"`

log N = – kt + log C

log N – log C = – kt

`log ("N"/"C")` = – kt

`"N"/"C"` = e^{– kt} 

N = `"Ce"^(-"kt")`  .........(1)

Initial condition:

When t = 0 we have N = N₀

N₀ = Ce^–k(0)

N₀ = Ce⁰

N₀ = C

Substituting C value in equation (1), we get,

N = N₀ e^–kt  ........(2)

Given: In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years be when t = p₀ 

N = `"N"_0 - 10/100  "N"_0 = "N"_0 - 1/10 "N"_0`

= `(10 "N"0 - "N"_0)/10`

= `(9"N"_0)/10`

∴ N = `(9"N"_0)/10`

 Substituting N =`(9"N"_0)/10` in equation (2), we get

`(9"N"_0)/10 ="N"_0  "e"^(-"kt")`

`9/10 = "e"^(-"k"(100))`

`log(9/10) = - "k" (100)`

∵ t = 100

`1/100 log(9/10) = - "k`

∴ k =`(-1)/100  log(9/10)`

Equation (2) becomes

N = `"N"_0 "e"^(+ 1/00 log (9/100) "r")`

`"N"/"N"_0 = "e"^(1/100 log(9/100)"t")`

`log("N"/"N"_0) = 1/100  log (9/10)"t"`  ........(3)

When t = 1000 equation (2) becomes

`log  "N"/"N"_0 = 1/100  log (9/10) xx 1000`

`log ("N"/"N"_0) = 1000/100 log (9/10)`

`log ("N"/"N"_0) = log(9/10)`

`"N"/"N"_0 = (9/100)^10`

The percentage of the original radioactive nuclei of remain after 1000 years is

`"N"/"N"_0 xx 100 = (9/10)^10 xx 100`

= `9^10/10^10 xx 10^2`

= `9^10/10^8`%