The half life a particular isotope of barium is 12s. What is the activity of a `1.0xx10^(-6)kg` sample of this isotope?

Options

A. `1.2xx10^(15)Bq`
B. `2.5xx10^(17)Bq`
C. `1.8xx10^(16)Bq`
D. `3.6xx10^(18)Bq`

Answer

To solve the problem of finding the activity of a `1.0 x 10^(-6) kg` sample of barium with a half-life of 12 seconds, we will follow these steps:

### Step 1: Calculate the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
where \(t_{1/2}\) is the half-life of the isotope.

Given:
- \(t_{1/2} = 12 \, \text{s}\)

Substituting the values:
\[ \lambda = \frac{\ln(2)}{12} \]

### Step 2: Calculate the number of atoms (N) in the sample
To find the number of atoms, we first need to calculate the number of moles in the sample using the molar mass of barium.

The molar mass of barium (Ba) is approximately \(137 \, \text{g/mol}\).

Convert the mass of the sample from kg to grams:
\[ 1.0 \times 10^{-6} \, \text{kg} = 1.0 \times 10^{-3} \, \text{g} \]

Now, calculate the number of moles (n):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.0 \times 10^{-3} \, \text{g}}{137 \, \text{g/mol}} \approx 7.30 \times 10^{-6} \, \text{mol} \]

Next, use Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{atoms/mol}\)) to find the number of atoms (N):
\[ N = n \times N_A = 7.30 \times 10^{-6} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 4.39 \times 10^{18} \, \text{atoms} \]

### Step 3: Calculate the activity (A)
The activity (A) can be calculated using the formula:
\[ A = \lambda N \]

Substituting the values we have:
\[ A = \left(\frac{\ln(2)}{12}\right) \times (4.39 \times 10^{18}) \]

Calculating this gives:
\[ A \approx 0.05776 \times 4.39 \times 10^{18} \approx 2.54 \times 10^{17} \, \text{disintegrations/second} \, (\text{or Bq}) \]

### Final Answer
The activity of the `1.0 x 10^(-6) kg` sample of barium is approximately:
\[ A \approx 2.54 \times 10^{17} \, \text{Bq} \]

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