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: Determine the number of moles of Radium-226
We know that the number of moles can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Given:
- Mass of Radium (Ra) = 1 g
- Molar mass of Radium-226 = 226 g/mol

Substituting the values:

\[ \text{Number of moles} = \frac{1 \, \text{g}}{226 \, \text{g/mol}} = \frac{1}{226} \, \text{mol} \]

### Step 2: Calculate the total number of atoms in 1 g of Radium-226
Using Avogadro's number (6.022 × 10²³ particles/mol), we can find the total number of atoms:

\[ \text{Total number of atoms} = \text{Number of moles} \times \text{Avogadro's number} \]

Substituting the values:

\[ \text{Total number of atoms} = \left(\frac{1}{226} \, \text{mol}\right) \times (6.022 \times 10^{23} \, \text{atoms/mol}) = \frac{6.022 \times 10^{23}}{226} \, \text{atoms} \]

Calculating this gives:

\[ \text{Total number of atoms} \approx 2.66 \times 10^{21} \, \text{atoms} \]

### Step 3: Use the formula for decay constant
The decay constant (λ) can be calculated using the formula:

\[ \lambda = \frac{\text{Number of disintegrations per second}}{\text{Total number of atoms present}} \]

Given that the number of alpha particles emitted per second (which is the number of disintegrations) is \(3.7 \times 10^{10}\):

Substituting the values:

\[ \lambda = \frac{3.7 \times 10^{10}}{2.66 \times 10^{21}} \]

Calculating this gives:

\[ \lambda \approx 1.39 \times 10^{-11} \, \text{s}^{-1} \]

### Final Answer
The decay constant (λ) for Radium-226 is approximately:

\[ \lambda \approx 1.39 \times 10^{-11} \, \text{s}^{-1} \]