Questions: Be sure to answer all parts. A radioactive substance undergoes decay as follows: Time (days) Mass (g) 0 500 1 389 2 303 3 236 4 184 5 143 6 112 Calculate the first-order decay constant and the half-life of the reaction. Constant: days Half-life: days

Be sure to answer all parts.
A radioactive substance undergoes decay as follows:
Time (days) Mass (g)
0 500
1 389
2 303
3 236
4 184
5 143
6 112

Calculate the first-order decay constant and the half-life of the reaction.
Constant: days
Half-life: days

Transcript text: Be sure to answer all parts. A radioactive substance undergoes decay as follows: \begin{tabular}{|c|c|} \hline Time (days) & Mass $(\mathbf{g})$ \\ \hline \hline 0 & 500 \\ \hline 1 & 389 \\ \hline \hline 2 & 303 \\ \hline 3 & 236 \\ \hline $\mathbf{4}$ & 184 \\ \hline $\mathbf{5}$ & 143 \\ \hline $\mathbf{6}$ & 112 \\ \hline \end{tabular} Calculate the first-order decay constant and the half-life of the reaction. Constant: $\square$ days $^{-1}$ Half-life: $\square$ days

Solution

Solution Steps

Step 1: Understanding the Decay Process

The decay of a radioactive substance can be modeled using the first-order decay equation: \[ N(t) = N_0 e^{-\lambda t} \] where:

$ N(t) $ is the mass at time $ t $,
$ N_0 $ is the initial mass,
$ \lambda $ is the decay constant,
$ t $ is the time.

Step 2: Setting Up the Equations

Given the data, we can use two points to find the decay constant $ \lambda $. Let's use the initial mass $ N_0 = 500 $ g at $ t = 0 $ and the mass $ N(1) = 389 $ g at $ t = 1 $ day.

Step 3: Solving for the Decay Constant

Using the first-order decay equation: \[ 389 = 500 e^{-\lambda \cdot 1} \] Solving for $ \lambda $: \[ \frac{389}{500} = e^{-\lambda} \] \[ \ln\left(\frac{389}{500}\right) = -\lambda \] \[ \lambda = -\ln\left(\frac{389}{500}\right) \] \[ \lambda = -\ln(0.778) \] \[ \lambda \approx 0.2513 \, \text{days}^{-1} \]

Step 4: Calculating the Half-Life

The half-life $ t_{1/2} $ of a first-order reaction is given by: \[ t_{1/2} = \frac{\ln(2)}{\lambda} \] Substituting the value of $ \lambda $: \[ t_{1/2} = \frac{\ln(2)}{0.2513} \] \[ t_{1/2} \approx 2.757 \, \text{days} \]