Question

The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.

Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.

`int dx/x = square + c` 

∴ logx = `square`

x = `square` = `square`.e^c

∴ x = `square`.a where a = e^c

At t = 0, x = 800

∴ a = `square`

At t = 5, x = 400

∴ e^–5k = `square`

Now when t = 30 

x = `square` × `square` = 800 × (e^–5k)⁶ = 800 × `square` = `square`.

The mass remaining after 30 days will be `square` mg.

Answer 1

Let x be the amount present at time t then

∴ `dx/dt` = –kx

 where k is constant of proportion

∴ `int dx/x = bb(-int kdt) + c` 

∴ logx = –kt + c

∴ x = e^–kt+c = e^–kt × e^c

∴ x = a.e^–kt where a = e^c

At t = 0, x = 800

∴ 800 = a · e⁰

∴ a = 800

∴ x = 800 · e^–kt

At t = 5, x = 400

∴ 400 = 800 · e^–5k

∴ e^–5k = `400/800 = 1/2`

∴ e^–5k = `bb(1/2)`

Now when t = 30,

x = 800 · e^–30k

= 800 · (e^–5k)⁶

= `800 xx (1/22)^6`

= 12.5

∴ The mass remaining after 30 days will be 12.5 mg.