Question

The disintegration rate of a radio-active sample is 10¹⁰ per hour at 20 hours from the start. It reduces to 5 × 10⁹ per hour after 30 hours. Calculate the decay constant.

Answer 1

Given:

A(t₁) = 10¹⁰ per hour, where t₁ = 20 h,

A(t₂) = 5 × 10⁹ per hour, where t₂ = 30 h,

To find: λ = ?

Formula:

A(t) = `A_0e^{-lambdat}`

∴ `(A(t_1))/(A(t_2)) = (A_0e^{-lambdat_1})/(A_0e^{-lambdat_2}) = (e^{-lambdat_1})/(e^{-lambdat_2})`

∴ `10^10/(5 xx 10^-9) = (e^{-lambda20})/(e^{-lambda30}) = e^{lambda(30 - 20)} = e^{10lambda}`

∴ 2 = `e^{10lambda}`

∴ 2 = `e^{10lambda}`

∴ log2 = 10λloge = 10λ ..............(∵ loge = 1)

∴ 0.693 = 10λ

∴ λ = 0.0693

Answer 2

`N = N_0e^-(lambdat)`

`N_2/N_1 = e^(-lambda(t_2-t_1)`

Given:

N₁​ = 1.0 × 10¹⁰ per hour at t₁ = 20 hours

N_2​ = 5.0 × 10⁹ per hour at t₂ = 30 hours

`N_2/N_1 = e^(-lambda(t_2-t_1)) =>(5xx10^9)/(1xx10^10) = e^(-lambda(10)) => 1/2 = e^(-10lambda)`

Take natural log

`(1/2) = -10lambda => -ln 2 = -10lambda => lambda = ln2/10 = 0.693/10`

= 0.0693 per hour