Answer :
We start with the formula for radioactive decay:
[tex]$$
m(t) = m_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}},
$$[/tex]
where:
- [tex]$m(t)$[/tex] is the remaining mass after time [tex]$t$[/tex],
- [tex]$m_0$[/tex] is the initial mass,
- [tex]$t$[/tex] is the elapsed time, and
- [tex]$T_{1/2}$[/tex] is the half-life of the substance.
Given:
- [tex]$m_0 = 100.0$[/tex] g,
- [tex]$t = 1600$[/tex] years, and
- [tex]$T_{1/2} = 5272$[/tex] years,
we substitute these values into the decay formula:
[tex]$$
m(1600) = 100.0 \left(\frac{1}{2}\right)^{\frac{1600}{5272}}.
$$[/tex]
1. First, calculate the exponent:
[tex]$$
\frac{1600}{5272} \approx 0.30349.
$$[/tex]
2. Next, evaluate the decay factor:
[tex]$$
\left(\frac{1}{2}\right)^{0.30349} \approx 0.81029.
$$[/tex]
3. Finally, compute the remaining mass:
[tex]$$
m(1600) = 100.0 \times 0.81029 \approx 81.03 \text{ g}.
$$[/tex]
Rounding to a reasonable number of significant figures, the remaining amount of cobalt-60 after 1600 years is approximately [tex]$81.1$[/tex] g.
Thus, the correct answer is:
[tex]$$\boxed{81.10 \text{ g}}.$$[/tex]
[tex]$$
m(t) = m_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}},
$$[/tex]
where:
- [tex]$m(t)$[/tex] is the remaining mass after time [tex]$t$[/tex],
- [tex]$m_0$[/tex] is the initial mass,
- [tex]$t$[/tex] is the elapsed time, and
- [tex]$T_{1/2}$[/tex] is the half-life of the substance.
Given:
- [tex]$m_0 = 100.0$[/tex] g,
- [tex]$t = 1600$[/tex] years, and
- [tex]$T_{1/2} = 5272$[/tex] years,
we substitute these values into the decay formula:
[tex]$$
m(1600) = 100.0 \left(\frac{1}{2}\right)^{\frac{1600}{5272}}.
$$[/tex]
1. First, calculate the exponent:
[tex]$$
\frac{1600}{5272} \approx 0.30349.
$$[/tex]
2. Next, evaluate the decay factor:
[tex]$$
\left(\frac{1}{2}\right)^{0.30349} \approx 0.81029.
$$[/tex]
3. Finally, compute the remaining mass:
[tex]$$
m(1600) = 100.0 \times 0.81029 \approx 81.03 \text{ g}.
$$[/tex]
Rounding to a reasonable number of significant figures, the remaining amount of cobalt-60 after 1600 years is approximately [tex]$81.1$[/tex] g.
Thus, the correct answer is:
[tex]$$\boxed{81.10 \text{ g}}.$$[/tex]
