What is the amount of `"^(60)CO_(27)` necessary to provide a radioactive source of 8.0 m Ci strength? The half-life of `^(60)CO_(27)` is 5.3 years.

Options

A. `7.12xx10^6 g`
B. `2.16 xx10^(-5) g`
C. `7.12 xx10^(-6) g`
D. `7.12 xx10^(-4) g`

Answer

To find the amount of \(^{60}\text{Co}_{27}\) necessary to provide a radioactive source of 8.0 mCi strength, we will follow these steps:

### Step 1: Convert the half-life from years to seconds
The half-life of \(^{60}\text{Co}_{27}\) is given as 5.3 years. We need to convert this into seconds.

\[ \text{Half-life} (T_{1/2}) = 5.3 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]

Calculating this gives:

\[ T_{1/2} = 5.3 \times 365 \times 24 \times 60 \times 60 \approx 1.67 \times 10^8 \text{ seconds} \]

### Step 2: Convert the activity from mCi to Bq
The activity is given as 8.0 mCi. We need to convert this into the SI unit of Becquerels (Bq).

1 mCi = \(3.7 \times 10^{10}\) Bq, hence:

\[ 8.0 \text{ mCi} = 8.0 \times 3.7 \times 10^{10} \text{ Bq} = 2.96 \times 10^{11} \text{ Bq} \]

### Step 3: Calculate the decay constant (\(\lambda\))
The decay constant \(\lambda\) can be calculated using the formula:

\[ \lambda = \frac{0.693}{T_{1/2}} \]

Substituting \(T_{1/2}\):

\[ \lambda = \frac{0.693}{1.67 \times 10^8 \text{ seconds}} \approx 4.14 \times 10^{-9} \text{ s}^{-1} \]

### Step 4: Calculate the number of radioactive nuclei (N)
Using the relationship between activity (A), decay constant (\(\lambda\)), and the number of radioactive nuclei (N):

\[ A = \lambda N \]

Rearranging gives:

\[ N = \frac{A}{\lambda} \]

Substituting the values:

\[ N = \frac{2.96 \times 10^{11} \text{ Bq}}{4.14 \times 10^{-9} \text{ s}^{-1}} \approx 7.15 \times 10^{19} \text{ nuclei} \]

### Step 5: Calculate the mass of \(^{60}\text{Co}_{27}\)
The molar mass of \(^{60}\text{Co}_{27}\) is approximately 60 g/mol. Using Avogadro's number (\(N_A = 6.022 \times 10^{23}\) atoms/mol), we can find the mass corresponding to \(N\) nuclei.

\[ \text{Mass} = N \times \frac{\text{Molar mass}}{N_A} \]

Substituting the values:

\[ \text{Mass} = 7.15 \times 10^{19} \times \frac{60 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 7.123 \times 10^{-6} \text{ g} \]

### Final Answer
The amount of \(^{60}\text{Co}_{27}\) necessary to provide a radioactive source of 8.0 mCi strength is approximately \(7.123 \times 10^{-6}\) grams.

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