Ten gram atoms of an alpha active element disintegrated in a sealed contain in one hour. the He gas collected is 11.2 cm3 at STP , calcualte the half - life of the radioactive isotope? .

Answer

To solve the problem of calculating the half-life of the radioactive isotope, we will follow these steps:

### Step 1: Determine the amount of helium gas produced
Given that 10 gram atoms of an alpha-active element disintegrated in one hour and produced 11.2 cm³ of helium gas at STP, we need to convert the volume of helium gas to moles.

**Calculation:**
At STP, 1 mole of gas occupies 22.4 liters (or 22400 cm³). Therefore, the number of moles of helium gas produced is:
\[ \text{Moles of He} = \frac{11.2 \, \text{cm}^3}{22400 \, \text{cm}^3/\text{mol}} = \frac{11.2}{22400} = 5 \times 10^{-4} \, \text{mol} \]

### Step 2: Relate the moles of helium to the disintegration of the alpha-active element
Each disintegration of the alpha-active element produces one helium atom (or one alpha particle). Therefore, the number of moles of the original element that disintegrated is equal to the moles of helium produced.

### Step 3: Calculate the initial amount of the radioactive element
The initial amount of the radioactive element (N₀) is given as 10 gram atoms. We need to convert this to moles:
\[ N_0 = 10 \, \text{gram atoms} = 10 \, \text{mol} \quad (\text{since 1 gram atom = 1 mol}) \]

### Step 4: Determine the remaining amount of the radioactive element after disintegration
The amount of the radioactive element that disintegrated is equal to the moles of helium produced, which is \(5 \times 10^{-4}\) mol. Thus, the remaining amount (N) is:
\[ N = N_0 - \text{Moles of He} = 10 - 5 \times 10^{-4} \approx 9.9995 \, \text{mol} \]

### Step 5: Use the radioactive decay formula to find the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{2.303}{T} \log \left(\frac{N_0}{N}\right) \]
Where:
- \(T = 1 \, \text{hour} = 3600 \, \text{seconds}\)
- \(N_0 = 10 \, \text{mol}\)
- \(N \approx 9.9995 \, \text{mol}\)

**Calculation:**
\[ \lambda = \frac{2.303}{3600} \log \left(\frac{10}{9.9995}\right) \]

Calculating the logarithm:
\[ \log \left(\frac{10}{9.9995}\right) \approx \log(1.00005) \approx 0.0000217 \]
Thus,
\[ \lambda \approx \frac{2.303}{3600} \times 0.0000217 \approx 5 \times 10^{-5} \, \text{h}^{-1} \]

### Step 6: Calculate the half-life (T₁/₂)
The half-life can be calculated using the formula:
\[ T_{1/2} = \frac{0.693}{\lambda} \]
Substituting the value of λ:
\[ T_{1/2} = \frac{0.693}{5 \times 10^{-5}} \approx 13857 \, \text{hours} \]

### Step 7: Convert hours to years
To convert hours to years:
\[ T_{1/2} \approx \frac{13857}{24 \times 365} \approx 1.582 \, \text{years} \]

### Final Answer
The half-life of the radioactive isotope is approximately **1.582 years**.

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