Question

Radon-222 3.8 days

How many half lives have occurred if 12.5% of Rn-222 remains?

I need someone to explain how you do this to me?

Answer 1

3 half-lives have occured

Further explanation

Given

Half-life Radon-222 = 3.8 days

12.5% of Rn-222 remains

Required

Half-lives occurred

Solution

General formulas used in decay:

`\large{\boxed{\bold{N_t=N_0((1)/(2))^{t/t(1)/(2) }}}`

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

12.5% of Rn-222 remains ⇒ Nt/No = 0.125

Input the value :

`\tt 0.125=(1)/(2)^(t/3.8)\\\\(1)/(2)^3=(1)/(2)^(t/3.8)\\\\3=t/3.8\\\\t=11.4\rightarrow 3~half-lives`