Question

Radium decomposes at the rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?

Answer 1

Let x be the amount of the radium at time t.

Then the rate of decomposition is `"dx"/"dt"` which is proportional to x.

∴ `"dx"/"dt" prop "x"`

∴ `"dx"/"dt"` = - kx, where k > 0

∴ `1/"x" "dx"` = - k dt

On integrating, we get

`int 1/"x" "dx" = - "k" int "dt"`

∴ log x = - kt + c

Let the original amount be x₀, i.e. x = x₀, when t = 0.

∴ log x₀ = - k × 0 + c     ∴ c = log x₀ 

∴ log x = - kt + log x₀ 

∴ log x - log x₀ = - kt

∴ `log ("x"/"x"_0)` = - kt       .....(1)

But p% of the amount disappears in one year,

∴ when t = 1, x= x₀ - p % of x₀ , i.e. x = x₀ - `"px"_0/100`

∴ `log (("x"_0 - "px"_0/100)/"x"_0) = - "k" xx 1`

∴ k = `- log (1 - "p"/100) = - log ((100 - p)/100)`

∴ (1) becomes, `log ("x"/"x"_0) = "t" log ((100 - "p")/100)`

When t = 2, then

`log ("x"/"x"_0) = 2 log ((100 - "p")/100) = log ((100 - "p")/100)^2`

∴ `"x"/"x"_0 = ((100 - "p")/100)^2`

∴ x = `((100 - "p")/100)^2 "x"_0 = (1 - "p"/100)^2 "x"_0`

∴ % left after 2 years = `(100 xx (1 - "p"/100)^3 "x"_0)/"x"_0`

∴ `= 100 (1 - "p"/100)^2 = [10 (1 - "p"/100)]^2`

`= (10 - "p"/10)^2`

Hence, `= (10 - "p"/10)^2`% of the amount will be left after 2 years.