Question

PLEASE HELP this is physical science.

Answer 1

Bone age : 22,920 years

Further explanation

Given

Nt = 2.5 g C-14

No = 40 g

half-life = 5730 years

Required

time of decay

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

Input the value :

`\tt 2.5=40.(1)/(2)^(t/5730)\\\\(2.5)/(40)=(1)/(2)^(t/5730)\\\\((1)/(2))^4=(1)/(2)^(t/5730)\\\\4=t/5730\rightarrow t=22920~years`