The problem is about finding the half-life of a radioactive material, given that it loses one-third of its mass in 2 minutes. This is a typical problem of exponential decay, and the half-life is the time required for the material to reduce to half of its original mass.

Step-by-Step Approach:

The formula for exponential decay is:

$M(t) = M_0 e^{-kt}$

Where:

• $M(t)$ is the mass at time $t$,
• $M_0$ is the initial mass,
• $k$ is the decay constant, and
• $t$ is time.

We know that after 2 minutes, one-third of the mass has decayed. Therefore, the remaining mass is two-thirds of the original mass. So,

$M(2) = \frac{2}{3} M_0$

Substitute this into the decay equation:

$\frac{2}{3} M_0 = M_0 e^{-2k}$

Now, cancel $M_0$ from both sides:

$\frac{2}{3} = e^{-2k}$

Take the natural logarithm on both sides:

$\ln\left(\frac{2}{3}\right) = -2k$

Now solve for $k$:

$k = -\frac{1}{2} \ln\left(\frac{2}{3}\right)$

Next, to find the half-life $t_{1/2}$, use the fact that when $M(t) = \frac{1}{2} M_0$:

$\frac{1}{2} M_0 = M_0 e^{-kt_{1/2}}$

Cancel $M_0$:

$\frac{1}{2} = e^{-kt_{1/2}}$

Take the natural logarithm on both sides:

$\ln\left(\frac{1}{2}\right) = -kt_{1/2}$

Solve for $t_{1/2}$:

$t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-k}$

Now substitute the value of $k$ found earlier to get the half-life. Let me calculate it for you.The half-life of the radioactive material is approximately 3.42 minutes.

Would you like further clarification on any step? Here are five related questions to help you explore this topic further:

1. How does the decay constant $k$ relate to the half-life in general?
2. What happens to the half-life if the decay rate increases?
3. How would the equation change if the material lost 50% of its mass in 2 minutes instead?
4. Can exponential decay be applied to non-radioactive processes? If so, how?
5. How does the exponential decay graph behave as time approaches infinity?

Tip: The half-life is independent of the initial quantity of the material; it only depends on the decay rate.