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

3 days

Step-by-step explanation:

Answer 2

3 days

Step-by-step 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 ((N_(0))/(N)\right ) & = & kt\\\\\ln\left ((100)/(10)\right ) & = & k* 33\\\\\ln10 & = & 33k\\2.303 & = & 33k\\k & = & \text{0.0698 da}^(-1)\\\end{array}`

(b) Calculate the half-life

`\begin{array}{rcl}t_{(1)/(2)} & = & (\ln 2)/(k )\\\\ & = & \frac{\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 ((N_(0))/(N)\right ) & = & kt\\\\\ln\left ((100)/(10)\right ) & = & k* 43\\\\\ln10 & = & 43k\\2.303 & = &43k\\k & = & \text{0.0535 da}^(-1)\\\end{array}`

(b) Calculate the half-life

`\begin{array}{rcl}t_{(1)/(2)} & = & (\ln 2)/(k )\\\\ & = & \frac{\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.