1 mixture of `Co^(58)` and `Co^(59)` has the activity of `2.2xx10^(12)` dis `s^(-1)` Half-life of `Co^(58)` is 71.3 days. Find percent by mass of `Co^(58)` in the mixture

Explanation

To solve the problem of finding the percent by mass of \( \text{Co}^{58} \) in a mixture of \( \text{Co}^{58} \) and \( \text{Co}^{59} \) with a given activity, we will follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the mass of \( \text{Co}^{58} \) in grams. Since the total mass of the mixture is 1 gram, the mass of \( \text{Co}^{59} \) will be \( 1 - x \) grams.

2. **Calculate Moles**:
The number of moles of \( \text{Co}^{58} \) can be calculated using the formula:
\[ n_{58} = \frac{x}{58} \]
The number of moles of \( \text{Co}^{59} \) is:
\[ n_{59} = \frac{1 - x}{59} \]

3. **Calculate Number of Atoms**:
Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \)), the number of atoms of \( \text{Co}^{58} \) is:
\[ N_{58} = n_{58} \times N_A = \frac{x}{58} \times 6.022 \times 10^{23} \]
The number of atoms of \( \text{Co}^{59} \) is:
\[ N_{59} = n_{59} \times N_A = \frac{1 - x}{59} \times 6.022 \times 10^{23} \]

4. **Calculate Decay Constant (\( \lambda \))**:
The decay constant \( \lambda \) is related to the half-life (\( t_{1/2} \)) by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Given that the half-life of \( \text{Co}^{58} \) is 71.3 days, we first convert this to seconds:
\[ t_{1/2} = 71.3 \times 24 \times 60 \times 60 \text{ seconds} \]
Then calculate \( \lambda \):
\[ \lambda = \frac{0.693}{71.3 \times 24 \times 60 \times 60} \]

5. **Calculate Activity**:
The activity \( A \) of \( \text{Co}^{58} \) is given by:
\[ A_{58} = \lambda N_{58} \]
Substituting \( N_{58} \):
\[ A_{58} = \lambda \left( \frac{x}{58} \times 6.022 \times 10^{23} \right) \]

6. **Total Activity**:
The total activity of the mixture is given as \( 2.2 \times 10^{12} \) dis/s. Since \( \text{Co}^{59} \) is stable, its contribution to activity is zero. Thus:
\[ A_{58} = 2.2 \times 10^{12} \]

7. **Set Up the Equation**:
Equating the expressions for activity:
\[ \lambda \left( \frac{x}{58} \times 6.022 \times 10^{23} \right) = 2.2 \times 10^{12} \]

8. **Solve for \( x \)**:
Rearranging gives:
\[ x = \frac{2.2 \times 10^{12} \times 58}{\lambda \times 6.022 \times 10^{23}} \]

9. **Calculate Percent by Mass**:
The percent by mass of \( \text{Co}^{58} \) is given by:
\[ \text{Percent by mass} = \left( \frac{x}{1} \right) \times 100 = x \times 100 \]

### Final Calculation:
After performing the calculations, we find that:
\[ x \approx 1.8 \times 10^{-3} \text{ grams} \]
Thus, the percent by mass of \( \text{Co}^{58} \) is approximately:
\[ \text{Percent by mass} \approx 0.18\% \]