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Approximate half-life
Radon-222 = 4 days

A sample contains 18mg of radon-222. You estimate that this is 0.01% of the original isotope. How many half-lives have passed?

Answer 1

13 half-lives have passed

Further explanation

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

t1/2 = 4 days

Nt=18 mg (0.01% of the original isotope)

18 mg (Nt) = 0.01% No

No = the original isotope :

`\tt No=(100)/(0.01)* 18~mg=180,000`

The duration of decay (T) :

`\tt (Nt)/(No)=(1)/(2)^(T/4)\rightarrow Nt=0.01\%No`

`\tt 0.01\%=(1)/(2)^(T/4)\\\\10^(-4)=(1)/(2)^(T/4)\\\\((1)/(2))^(13)=(1)/(2)^(T/4)\\\\13=T/4\rightarrow T=52~days`

Half-lives passed :

`\tt (52)/(4)=13~half-lives`