A radioactive material loses 34% of its mass in 8 minutes. what is its half-life?

Neetesh Kumar | September 05, 2024

This is the solution to Myopenmath Math2A Differential Equation
** Growth and Decay** homework question number 4.1.3
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Step-by-Step-Solution:
We are given that a radioactive material loses 34% of its mass in 8 minutes. We need to find its half-life.

The exponential decay model is given by the equation:

$Q(t) = Q_0 e^{-kt}$

Where:

• $Q(t)$ is the quantity of material remaining after time $t$,
• $Q_0$ is the initial quantity of the material,
• $k$ is the decay constant,
• $t$ is the time.

Step 1: Relating the Given Information

We are told that 34% of the material is lost, meaning 66% remains. This can be expressed as:

$Q(t) = 0.66 Q_0$

We are also told that this occurs at $t = 8$ minutes. Substituting this into the exponential decay equation:

$0.66 Q_0 = Q_0 e^{-k \cdot 8}$

Step 2: Simplifying the Equation

Cancel out $Q_0$ from both sides:

$0.66 = e^{-8k}$

Step 3: Taking the Natural Logarithm of Both Sides

Take the natural logarithm of both sides to solve for $k$:

$\ln(0.66) = -8k$

Step 4: Solving for the Decay Constant $k$

Now solve for $k$:

$k = \frac{-\ln(0.66)}{8} \approx 0.05226$

Step 5: Finding the Half-Life

The half-life $t_{\frac{1}{2}}$ is related to the decay constant $k$ by the equation:

$k = \frac{\ln(2)}{t_{\frac{1}{2}}}$

Solving for the half-life $t_{\frac{1}{2}}$:

$t_{\frac{1}{2}} = \frac{\ln(2)}{k}$

Substitute the value of $k$:

$t_{\frac{1}{2}} = \frac{\ln(2)}{0.05226} \approx 13.35 \, \text{minutes}$

Thus, the half-life of the material is

approximately $13.35$ minutes.

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