A small bottle contains powered beryllium Be & gaseous radon which is used as a source of `alpha`-particles. Netrons are produced when `alpha-`particles of the radon react with beryllium. The yield of this reaction is `(1//4000)` i.e. only one `alpha`-particle out of induced the reaction. Find the amount of radon `(Rn^(222))` originally introduced into the source, if it prouduces `1.2 xx 10^(6)` neutrons per second after `7.6` days. [`T_(1//2)` of `R_(n) = 3.8` days]

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We need to find the initial amount of radon (Rn-222) that produces 1.2 x 10^6 neutrons per second after 7.6 days, given that the yield of the reaction is 1/4000 and the half-life of radon is 3.8 days.

### Step 2: Determine the number of half-lives
The time given is 7.6 days. Since the half-life of radon is 3.8 days, we can determine how many half-lives have passed:
\[ \text{Number of half-lives} = \frac{7.6 \text{ days}}{3.8 \text{ days}} = 2 \]
This means that after 7.6 days, the radon has undergone 2 half-lives.

### Step 3: Calculate the remaining amount of radon
If we start with an initial amount \( N_0 \), after 2 half-lives, the remaining amount of radon is:
\[ N = \frac{N_0}{2^2} = \frac{N_0}{4} \]

### Step 4: Calculate the decay constant
The decay constant \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{0.693}{3.8 \text{ days}} = \frac{0.693}{3.8 \times 24 \times 3600 \text{ seconds}} \approx 2.11 \times 10^{-6} \text{ s}^{-1} \]

### Step 5: Relate activity to the number of nuclei
The activity \( A \) is given by:
\[ A = N \lambda \]
Substituting \( N = \frac{N_0}{4} \):
\[ A = \frac{N_0}{4} \lambda \]
Given that the activity corresponds to the number of neutrons produced per second:
\[ 1.2 \times 10^6 = \frac{N_0}{4} \cdot 2.11 \times 10^{-6} \]

### Step 6: Solve for \( N_0 \)
Rearranging the equation to solve for \( N_0 \):
\[ N_0 = 4 \cdot 1.2 \times 10^6 \cdot \frac{1}{2.11 \times 10^{-6}} \]
Calculating \( N_0 \):
\[ N_0 \approx 4 \cdot 1.2 \times 10^6 \cdot 4.74 \times 10^5 \approx 9.095 \times 10^{15} \text{ nuclei} \]

### Step 7: Convert to mass
To find the mass of radon, we use the molar mass of radon (approximately 222 g/mol):
\[ \text{Mass} = N_0 \cdot \frac{\text{Molar mass}}{N_A} \]
Where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \text{ nuclei/mol} \)):
\[ \text{Mass} = 9.095 \times 10^{15} \cdot \frac{222 \text{ g/mol}}{6.022 \times 10^{23} \text{ nuclei/mol}} \approx 3.354 \times 10^{-4} \text{ g} \]

### Final Answer
The initial amount of radon introduced into the source is approximately \( 3.354 \times 10^{-4} \) grams.

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