Respuesta :
The amount of substance at any time t is given by the equation,
At = A1 x e^-kt
From the given half-life,
At/A1 = 0.5 = e^-k(1599)
The value of k is 4.3349x10^-4
Using the same equation for the next items
(1000 years) At = (250 g) x e^(-4.3349x10^-4)(1000) = 162.06 grams
(2000 years) At = (250 g) x e^(-4.3349x10^-4)(2000) = 105.05 grams
At = A1 x e^-kt
From the given half-life,
At/A1 = 0.5 = e^-k(1599)
The value of k is 4.3349x10^-4
Using the same equation for the next items
(1000 years) At = (250 g) x e^(-4.3349x10^-4)(1000) = 162.06 grams
(2000 years) At = (250 g) x e^(-4.3349x10^-4)(2000) = 105.05 grams
Answer: The amount of Ra-226 isotope after 1000 years is 162.14 grams and after 2000 years are 105.16 grams
Explanation:
The equation used to calculate half life for first order kinetics:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
where,
k = rate constant of the reaction = ?
[tex]t_{1/2}[/tex] = half life of the reaction = 1599 years
Putting values in above equation, we get:
[tex]k=\frac{0.693}{1599yrs}=4.33\times 10^{-4}yr^{-1}[/tex]
Integrated rate law expression for first order kinetics is given by the equation:
[tex]N=N_oe^{-kt}[/tex]
where,
N = amount left after time 't'
[tex]N_o[/tex] = initial amount = 250 grams
t = time taken =
k = rate constant = [tex]4.33\times 10^{-4}yr^{-1}[/tex]
- When t = 1000 years
Putting values in above equation, we get:
[tex]N=250\times e^{-(4.33\times 10^{-4}\times 1000)}\\\\N=162.14g[/tex]
- When t = 2000 years
Putting values in above equation, we get:
[tex]N=250\times e^{-(4.33\times 10^{-4}\times 2000)}\\\\N=105.16g[/tex]
Hence, the amount of Ra-226 isotope after 1000 years is 162.14 grams and after 2000 years are 105.16 grams