Fluorine-21 has a half life of approximately 5 seconds. what fraction of the original nuclei would remain after 1 minute?

We can use two equations for this problem.

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

5 s = 0.693 / λ 
λ   = 0.693 / 5 s        (1) 

Nt = Nο eΛ(-λt)
Nt/No =  eΛ(-λt)        (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 1 minute  = 60 s

From (1) and (2),
Nt/No =  eΛ(-(0.693 / 5 s) 60 s)
Nt/No = 2.445 x 10⁻⁴

Hence, the fraction of the original nuclei would remain after 1 minute is 2.445 x 10⁻⁴.

The answer is b on plato.

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