Question

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isotope B does this in 43 days. What is the approximate difference in the half-lives of the isotopes? 3 days 10 days 13 days 33 days

Answer 1

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

$\begin{array}{rcl}\ln\left (\dfrac{N_{0}}{N}\right ) & = & kt\\\\\ln\left (\dfrac{100}{10}\right ) & = & k\times 33\\\\\ln10 & = & 33k\\2.303 & = & 33k\\k & = & \text{0.0698 da}^{-1}\\\end{array}$

(b) Calculate the half-life

$\begin{array}{rcl}t_{\frac{1}{2}} & = & \dfrac{\ln 2}{k }\\\\ & = & \dfrac{\ln 2}{ \text{0.0698 da}^{-1} }\\\\& = & \textbf{9.9 da}\\\end{array}$

2. Isotope B

Data:

N₀ = 100 %

N =    10 %

 t =    43 da

(a) Calculate the value of k

$\begin{array}{rcl}\ln\left (\dfrac{N_{0}}{N}\right ) & = & kt\\\\\ln\left (\dfrac{100}{10}\right ) & = & k\times 43\\\\\ln10 & = & 43k\\2.303 & = &43k\\k & = & \text{0.0535 da}^{-1}\\\end{array}$

(b) Calculate the half-life

$\begin{array}{rcl}t_{\frac{1}{2}} & = & \dfrac{\ln 2}{k }\\\\ & = & \dfrac{\ln 2}{ \text{0.0535 da}^{-1} } \\\\& = & \textbf{12.9 da}\\\end{array}$

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.