Answer:
t = 5.59x10⁴ y
Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
[tex] A_{t} = A_{0}\cdot e^{- \lambda t} [/tex] (1)
where [tex]A_{t}[/tex]: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.
To find A₀ we can use the following equation:
[tex] A_{0} = N_{0} \lambda [/tex] (2)
where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
[tex] N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}} [/tex]
where [tex]m_{T}[/tex]: is the tree's carbon mass, [tex]N_{A}[/tex]: is the Avogadro's number and [tex]m_{^{12}C}[/tex]: is the ¹²C mass.
[tex] N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C [/tex]
Similarly, from equation (2) λ is:
[tex] \lambda = \frac{Ln(2)}{t_{1/2}} [/tex]
where t 1/2: is the half-life of ¹⁴C= 5700 years
[tex] \lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1} [/tex]
So, the initial activity A₀ is:
[tex] A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y [/tex]
Finally, we can calculate the time from equation (1):
[tex] t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y [/tex]
I hope it helps you!
