Answer:
8923 years
Step-by-step explanation:
Half life of C-14 = 5730yrs
decay rate= 34%
Halt life (t^1/2) = (ln2) / k
5730 = (ln2) /k
k = (ln2) / 5730
k = 1.209 * 10^-4
For first order reaction in radioactivity,
ln(initial amount) = -kt
ln(34/100) = -(1.209*10^-4)t
-1.0788 = -(1.209*10^-4)t
t = -1.7088/ -1.209*10^-4
t = 8923 years
It would take 8918 years for the tree to decay to 34%.
The half life is the time taken for a substance to decay to half of its value. It is given by:
[tex]N(t) = N_0(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} } \\\\where\ t=period, N(t)=value\ after\ t\ years, N_o=original\ amount, t_\frac{1}{2} =half\ life\\\\Given\ t_\frac{1}{2} =5730,N(t)=0.34N_o, hence:\\\\0.34N_o=N_o(\frac{1}{2} )^\frac{t}{5730} \\\\t=8918\ years[/tex]
It would take 8918 years for the tree to decay to 34%.
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