Answer:
[tex]3688.323[/tex]years
Explanation:
Given-
Half life of [tex]14[/tex]C [tex]= 5730[/tex]years
As we know -
[tex]A_{(t)} = A_0e^{kt}[/tex]
Where
[tex]A_{(t)} =[/tex] Mass of radioactive carbon after a time period "t"
[tex]A_0=[/tex] initial mass of radioactive carbon
[tex]k =[/tex]radioactive decay constant
[tex]t =[/tex]time
First we will find the value of "k"
[tex]\frac{1}{2} = (1)*e^{k*5730}\\[/tex]
On solving, we get -
[tex]e^{5730*k}= 0.5\\5730*k = ln(0.5)\\k = -0.000121[/tex]
Now, when mass of 14C becomes [tex]64[/tex]% of the plant material on earth today, then its age would be
[tex]A_{(t)} = A_0*e^{(-0.000121*t)}\\A_{(t)}= 0.64*A_0\\0.64*A_0 = A_0*e-^{(0.000121*t)}\\t = 3688.323[/tex]years