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:  

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

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 :

$\tt 2.5=40.\dfrac{1}{2}^{t/5730}\\\\\dfrac{2.5}{40}=\dfrac{1}{2}^{t/5730}\\\\(\dfrac{1}{2})^4=\dfrac{1}{2}^{t/5730}\\\\4=t/5730\rightarrow t=22920~years$