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:
[tex]\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}[/tex]
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 :
[tex]\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[/tex]
