A radioactive substance has a half-life of 80 days. If you start with 6.0 x 1010 atoms, how many atoms will remain after 320 days?

The number of half-lives is the total time divided by the half-life period.
Given:

• Half-life ($T_{1/2}$) = 80 days
• Total time ($t$) = 320 days

$$n = \frac{t}{T_{1/2}}$$ $$n = \frac{320 \, \text{days}}{80 \, \text{days}}$$ $$n = 4$$

So, 4 half-lives have passed.

The formula for radioactive decay is:

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

Where:

• $N_0$ is the initial number of atoms
• $N$ is the number of atoms remaining after time $t$
• $n$ is the number of half-lives

Given:

• Initial number of atoms ($N_0$) = $6.0 \times 10^{10}$
• Number of half-lives ($n$) = 4

Substitute the values into the formula:

$$N = (6.0 \times 10^{10}) \left(\frac{1}{2}\right)^4$$ $$N = (6.0 \times 10^{10}) \left(\frac{1}{16}\right)$$ $$N = \frac{6.0 \times 10^{10}}{16}$$ $$N = 0.375 \times 10^{10}$$

To express this in standard scientific notation, we adjust the decimal:

$$N = 3.75 \times 10^9$$

The calculated number of atoms remaining is $3.75 \times 10^9$.
This matches option A.

The number of atoms that would remain at the end of 320 days is \boxed{\text{3.75 \times 10^9}}.