Question:

A radioactive substance decays exponentially over time according to the equation $P(t) = P_0 e^{kt}$, where $P(t)$ is the amount of the substance at time $t$, $P_0$ is the initial amount of the substance, and $k$ is the decay constant.

Suppose that a sample of the substance starts with an initial amount of $500$ grams and decays at a rate of $0.05$ grams per second. Find the time it takes for the sample to decay to $100$ grams, rounded to the nearest hundredth of a second.

Answer:

Let's first write the given equation, $P(t) = P_0 e^{kt}$, in terms of the given information.

Given:

• $P_0 = 500$ grams (initial amount)
• Rate of decay = $0.05$ grams per second

To find the decay constant $k$, we use the fact that the rate of decay is equal to $kP(t)$:

$\frac{dP}{dt} = kP(t)$

Substituting $P(t) = P_0 e^{kt}$, we get:

$\frac{dP}{dt} = k \cdot 500e^{kt}$

Given that the rate of decay is $0.05$ grams per second, we have:

$0.05 = k \cdot 500e^{kt}$

Next, we solve for $k$:

$k \cdot 500e^{kt} = 0.05$

$e^{kt} = \frac{0.05}{500k} = \frac{1}{10000k}$

Taking the natural logarithm (ln) of both sides:

$kt = \ln\left(\frac{1}{10000k}\right)$

Now, we can solve for $t$:

$t = \frac{\ln\left(\frac{1}{10000k}\right)}{k}$

Given that the sample decays to $100$ grams, we can substitute $P(t) = 100$ into the equation $P(t) = P_0 e^{kt}$ and solve for $t$:

$100 = 500e^{kt}$

$e^{kt} = \frac{100}{500} = \frac{1}{5}$

Taking the natural logarithm (ln) of both sides:

$kt = \ln\left(\frac{1}{5}\right) = -\ln(5)$

Now, we can substitute this value of $kt$ into the previous equation to find $t$:

$t = \frac{\ln\left(\frac{1}{5}\right)}{k}$

Using a calculator, we can substitute the value of $k$ and evaluate $t$:

$t = \frac{\ln\left(\frac{1}{5}\right)}{-0.05}$

$t \approx 27.726$ seconds

Therefore, it takes approximately $27.73$ seconds (rounded to the nearest hundredth) for the sample to decay to $100$ grams.