A radioactive substance has an initial mass of 64 g and decays to 2 g in 24 days. What is its half-life?

From the image, we have:

• Total time elapsed ($t$) = 24 days
• Initial mass ($N_0$) = 64 g
• Final mass ($N$) = 2 g
• We need to find the half-life ($T_{1/2}$).

The formula for radioactive decay is:

$$N = N_0 \left(\frac{1}{2}\right)^n$$

Substitute the given values:

$$2 \text{ g} = 64 \text{ g} \left(\frac{1}{2}\right)^n$$

Divide both sides by 64 g:

$$\frac{2}{64} = \left(\frac{1}{2}\right)^n$$ $$\frac{1}{32} = \left(\frac{1}{2}\right)^n$$

Since $2^5 = 32$, we can write $\frac{1}{32}$ as $\left(\frac{1}{2}\right)^5$.

$$\left(\frac{1}{2}\right)^5 = \left(\frac{1}{2}\right)^n$$

Therefore, the number of half-lives ($n$) is 5.

The total time elapsed ($t$) is the product of the number of half-lives ($n$) and the half-life period ($T_{1/2}$):

$$t = n \times T_{1/2}$$

Substitute the known values:

$$24 \text{ days} = 5 \times T_{1/2}$$

Now, solve for $T_{1/2}$:

$$T_{1/2} = \frac{24 \text{ days}}{5}$$ $$T_{1/2} = 4.8 \text{ days}$$

The final answer is $\text{4.8 days}$.