Answer:
3 days
Explanation:
For each isotope, we must use the integrated rate law for a first order reaction to get the rate constant k, then use that to calculate the half-life t_½.
1. Isotope A
The integrated rate law for a first-order decay is
ln(N₀/N) = kt
Data:
N₀ = 100 %
N = 10 %
t = 33 da
(a) Calculate the value of k
[tex]\begin{array}{rcl}\ln\left (\dfrac{N_{0}}{N}\right ) & = & kt\\\\\ln\left (\dfrac{100}{10}\right ) & = & k\times 33\\\\\ln10 & = & 33k\\2.303 & = & 33k\\k & = & \text{0.0698 da}^{-1}\\\end{array}[/tex]
(b) Calculate the half-life
[tex]\begin{array}{rcl}t_{\frac{1}{2}} & = & \dfrac{\ln 2}{k }\\\\ & = & \dfrac{\ln 2}{ \text{0.0698 da}^{-1} }\\\\& = & \textbf{9.9 da}\\\end{array}[/tex]
2. Isotope B
Data:
N₀ = 100 %
N = 10 %
t = 43 da
(a) Calculate the value of k
[tex]\begin{array}{rcl}\ln\left (\dfrac{N_{0}}{N}\right ) & = & kt\\\\\ln\left (\dfrac{100}{10}\right ) & = & k\times 43\\\\\ln10 & = & 43k\\2.303 & = &43k\\k & = & \text{0.0535 da}^{-1}\\\end{array}[/tex]
(b) Calculate the half-life
[tex]\begin{array}{rcl}t_{\frac{1}{2}} & = & \dfrac{\ln 2}{k }\\\\ & = & \dfrac{\ln 2}{ \text{0.0535 da}^{-1} } \\\\& = & \textbf{12.9 da}\\\end{array}[/tex]
3. Calculate the difference in half-lives
Difference = 12.9 da - 9.9 da ≈ 3 da
The graphs below show the decay curves for isotopes A and B. The half-lives are approximately 10 da and 13 da, a difference of 3 da.
