The number of `alpha` particles emitted per second by 1g of `88^226` Ra is `3.7 times 10^10`. The decay constant is :

Options

A. `1.39 times 10^-11 sec^-1`
B. `13.9 times 10^-11 sec^-1`
C. `139 times 10^-10 sec^-1`
D. `13.9 times 10^-10 sec^-1`

Answer

To find the decay constant (λ) for the given problem, we can follow these steps:

### Step 1: Understand the relationship between decay constant, activity, and number of atoms.
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{A}{N} \]
where:
- \(A\) is the activity (number of disintegrations per second, which is given as \(3.7 \times 10^{10}\) alpha particles per second),
- \(N\) is the total number of radioactive atoms present.

### Step 2: Calculate the number of moles of \(^{226}_{88}Ra\).
To calculate the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given that we have 1 g of \(^{226}_{88}Ra\) and its molar mass is approximately 226 g/mol:
\[ \text{Number of moles} = \frac{1 \text{ g}}{226 \text{ g/mol}} = \frac{1}{226} \text{ mol} \]

### Step 3: Calculate the total number of atoms (N).
Using Avogadro's number (\(N_A = 6.022 \times 10^{23} \text{ atoms/mol}\)), we can find the total number of atoms:
\[ N = \text{Number of moles} \times N_A = \left(\frac{1}{226}\right) \times (6.022 \times 10^{23}) \text{ atoms} \]
Calculating this gives:
\[ N = \frac{6.022 \times 10^{23}}{226} \approx 2.66 \times 10^{21} \text{ atoms} \]

### Step 4: Substitute values into the decay constant formula.
Now we can substitute \(A\) and \(N\) into the decay constant formula:
\[ \lambda = \frac{3.7 \times 10^{10}}{2.66 \times 10^{21}} \]

### Step 5: Calculate the decay constant (λ).
Performing the division:
\[ \lambda \approx 1.39 \times 10^{-11} \text{ s}^{-1} \]

### Final Answer:
The decay constant \(λ\) is approximately:
\[ \lambda \approx 1.39 \times 10^{-11} \text{ s}^{-1} \]

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