Obtain the amount of `^60Co` necessary to provide a radioactive source of 8.0 Ci strength. The half-life of `^60Co` is 5.3 years?

Explanation

To solve the problem of determining the amount of \(^{60}Co\) necessary to provide a radioactive source of 8.0 Ci strength, we can follow these steps:

### Step 1: Convert Curie to Disintegrations per Second
The first step is to convert the activity from Curie (Ci) to disintegrations per second (dps). The conversion factor is:
\[ 1 \text{ Ci} = 3.7 \times 10^{10} \text{ dps} \]
Thus, for 8.0 Ci:
\[ R = 8.0 \text{ Ci} \times 3.7 \times 10^{10} \text{ dps/Ci} = 2.96 \times 10^{11} \text{ dps} \]

### Step 2: Calculate the Decay Constant (\( \lambda \))
The decay constant \( \lambda \) can be calculated using the half-life (\( T_{1/2} \)) formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Given that the half-life of \(^{60}Co\) is 5.3 years, we first convert this to seconds:
\[ T_{1/2} = 5.3 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \approx 1.67 \times 10^8 \text{ seconds} \]
Now substituting this into the decay constant formula:
\[ \lambda = \frac{0.693}{1.67 \times 10^8 \text{ s}} \approx 4.14 \times 10^{-9} \text{ s}^{-1} \]

### Step 3: Calculate the Number of Nuclei (\( N \))
Using the relationship between activity \( R \), decay constant \( \lambda \), and the number of nuclei \( N \):
\[ R = N \lambda \implies N = \frac{R}{\lambda} \]
Substituting the values we have:
\[ N = \frac{2.96 \times 10^{11} \text{ dps}}{4.14 \times 10^{-9} \text{ s}^{-1}} \approx 7.14 \times 10^{19} \text{ nuclei} \]

### Step 4: Calculate the Mass of \(^{60}Co\)
To find the mass, we use the formula:
\[ \text{mass} = \frac{N}{N_A} \times M \]
where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \text{ nuclei/mol} \)) and \( M \) is the molar mass of \(^{60}Co\) (approximately 60 g/mol).
Substituting the values:
\[ \text{mass} = \frac{7.14 \times 10^{19} \text{ nuclei}}{6.022 \times 10^{23} \text{ nuclei/mol}} \times 60 \text{ g/mol} \]
Calculating this gives:
\[ \text{mass} \approx 7.14 \times 60 \times 10^{-4} \text{ g} \approx 7.11 \times 10^{-3} \text{ g} \]

### Final Answer
The amount of \(^{60}Co\) necessary to provide a radioactive source of 8.0 Ci strength is approximately \( 7.11 \times 10^{-3} \) grams.

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