Answer:
t = 3,928 10⁵ years
Explanation:
For this exercise we must use the expression for the rate of radioactive decay
N =
Let's start by finding the amount of ¹⁴C in a sample of mass m = 0.100 kg. Let's use a three of direct proportions, if a mole of carbon has a mass of 12.01 g /mol and this has avogadro's number (6.022 10²³) of atoms, how many atoms are there in m = 0.1 kg
#_atoms = 0.1 (6.022 10²³ / 12.01) = 5.01 10²¹ carbon atoms
A carbon mixture has a 1.3 10-12 fraction ion of ¹⁴C atoms (N₀)
N₀ = #_atoms fraction
N₀ = 5.01 10²¹ 1.3 10⁻¹²
N₀ = 6,513 10²¹ radioactive ¹⁴C atoms
the activity constant (λ) can be found from the mean life time, = 5730 years
let's reduce to the SI system
T_{\frac{1}{2} }= 5730 year (365 days/ 1 year) (24 h/ 1 day) (3600 s/ 1 h)
T_{\frac{1}{2} }= 1.807 10¹¹
λ=
λ = 3.836 10⁻¹² decays / s
we substitute in the initial expression
N = 6.513 10²¹
in the exercise they indicate that 15 decays / s is counted
ln N/N₀ = - 3,836 10⁻¹² t
t =
we calculate
t = - (3, 836 10⁻¹²)⁻¹ ln ( )
t = 1,239 10¹³ s
let's reduce to years
t = 1,239 10¹³ s (1h / 3600 s) (1 day / 24h) (1 year / 365 day)
t = 3,928 10⁵ years