The decay constant of `Ra^(226)` is `1.37xx10^(-11)s^(-1)`. A sample of `Ra^(226)` having an activity of `1.5` millicurie will contain

Options

A. `4.05xx10^(18)` atoms
B. `3.7xx10^(17)` atoms
C. `2.05xx10^(15)` atoms
D. `4.7 xx 10^(10)` atoms

Explanation

To solve the problem, we need to find the number of atoms in a sample of \( Ra^{226} \) given its activity and decay constant. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between activity, decay constant, and number of atoms
The activity \( A \) of a radioactive sample is related to the decay constant \( \lambda \) and the number of radioactive atoms \( N \) by the formula:
\[ A = \lambda N \]
Where:
- \( A \) is the activity in disintegrations per second (Becquerels),
- \( \lambda \) is the decay constant in \( s^{-1} \),
- \( N \) is the number of radioactive atoms.

### Step 2: Convert activity from millicuries to disintegrations per second
Given that the activity of the sample is \( 1.5 \) millicuries, we need to convert this to disintegrations per second. We know that:
\[ 1 \text{ millicurie} = 3.7 \times 10^7 \text{ disintegrations per second} \]
Thus, for \( 1.5 \) millicuries:
\[ A = 1.5 \times 3.7 \times 10^7 = 5.55 \times 10^7 \text{ disintegrations per second} \]

### Step 3: Rearrange the activity formula to find the number of atoms
Now, we can rearrange the activity formula to solve for \( N \):
\[ N = \frac{A}{\lambda} \]

### Step 4: Substitute the values into the formula
We have:
- \( A = 5.55 \times 10^7 \) disintegrations per second,
- \( \lambda = 1.37 \times 10^{-11} \, s^{-1} \).

Substituting these values into the equation gives:
\[ N = \frac{5.55 \times 10^7}{1.37 \times 10^{-11}} \]

### Step 5: Calculate the number of atoms
Now, performing the calculation:
\[ N = \frac{5.55 \times 10^7}{1.37 \times 10^{-11}} \approx 4.05 \times 10^{18} \text{ atoms} \]

### Final Answer
Thus, the sample of \( Ra^{226} \) contains approximately \( 4.05 \times 10^{18} \) atoms.

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