Question

How many milligrams of a 6.5×10-2kg sample of the x parent isotope will remain at the end of 13.5 billion years at the half-life of three.

Answer 1

To determine how much of a parent isotope remains after a specific amount of time has passed, we can use the formula for radioactive decay:

\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]

Where:
- \(N\) = remaining quantity of the isotope
- \(N_0\) = initial quantity of the isotope
- \(t\) = time elapsed
- \(T_{1/2}\) = half-life period

### Given:
- Initial mass \(N_0 = 6.5 \times 10^{-2} \text{ kg} = 65 \text{ grams}\) (since \(1 \text{ kg} = 1000 \text{ grams}\))
- Half-life \(T_{1/2} = 3 \text{ billion years} = 3 \times 10^9 \text{ years}\)
- Time elapsed \(t = 13.5 \text{ billion years} = 13.5 \times 10^9 \text{ years}\)

### Step 1: Calculate the number of half-lives that have passed.
\[
\text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{13.5 \times 10^9 \text{ years}}{3 \times 10^9 \text{ years}} = 4.5
\]

### Step 2: Calculate the remaining quantity of the isotope.
Using the decay formula:

\[
N = 65 \text{ grams} \left( \frac{1}{2} \right)^{4.5}
\]

To compute \(\left( \frac{1}{2} \right)^{4.5}\):

\[
\left( \frac{1}{2} \right)^{4.5} = \frac{1}{\sqrt{2^9}} = \frac{1}{\sqrt{512}} \approx \frac{1}{22.6274} \approx 0.0442
\]

Now calculate \(N\):

\[
N \approx 65 \times 0.0442 \approx 2.873 \text{ grams}
\]

### Step 3: Convert grams to milligrams.
To convert grams to milligrams:

\[
N \approx 2.873 \text{ grams} \times 1000 \text{ mg/g} \approx 2873 \text{ mg}
\]

Therefore, at the end of 13.5 billion years, approximately **2873 milligrams** of the parent isotope will remain.