10.0 gram-atom of an `alpha`-active radioisotope is disintegrating in a sealed container. In one hour, the He gas collected at STP is 11.2 litres. Calculate the half-life of the isotope supposing each nucleus yielding one He atom.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the number of atoms in 10 gram-atoms of the radioisotope.
1. **Given**: 10 gram-atoms of the radioisotope.
2. **Moles**: Since 1 gram-atom is equivalent to 1 mole, 10 gram-atoms equal 10 moles.
3. **Avogadro's Number**: The number of atoms in 10 moles can be calculated using Avogadro's number, which is approximately \(6.022 \times 10^{23}\) atoms/mole.
\[ \text{Number of atoms} = 10 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 6.022 \times 10^{24} \, \text{atoms} \]

### Step 2: Calculate the number of helium atoms produced.
1. **Volume of Helium Collected**: 11.2 liters of helium gas is collected at STP (Standard Temperature and Pressure).
2. **Volume Conversion**: Convert liters to cubic centimeters (1 liter = 1000 cm³):
\[ 11.2 \, \text{liters} = 11.2 \times 1000 = 11200 \, \text{cm}^3 \]
3. **Molar Volume at STP**: The molar volume of a gas at STP is 22.4 liters (or 22400 cm³).
4. **Calculate Moles of Helium**:
\[ \text{Moles of He} = \frac{11200 \, \text{cm}^3}{22400 \, \text{cm}^3/\text{mole}} = 0.5 \, \text{moles} \]
5. **Calculate Number of Helium Atoms**:
\[ \text{Number of He atoms} = 0.5 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 3.011 \times 10^{23} \, \text{atoms} \]

### Step 3: Relate the number of disintegrated radioisotope atoms to helium atoms.
1. **Disintegration**: Each disintegration of a radioisotope nucleus produces one helium atom. Therefore, the number of disintegrated radioisotope atoms is equal to the number of helium atoms produced:
\[ N_0 - N = 3.011 \times 10^{23} \]
where \(N_0\) is the initial number of radioisotope atoms and \(N\) is the remaining number of radioisotope atoms after one hour.

### Step 4: Calculate the remaining number of radioisotope atoms.
1. **Initial Number of Atoms**: \(N_0 = 6.022 \times 10^{24}\)
2. **Remaining Atoms**:
\[ N = N_0 - 3.011 \times 10^{23} = 6.022 \times 10^{24} - 3.011 \times 10^{23} = 5.721 \times 10^{24} \]

### Step 5: Use the decay constant formula to find the half-life.
1. **Decay Constant (\(\lambda\))**: The decay constant can be calculated using the formula:
\[ \lambda = \frac{2.303}{t} \log\left(\frac{N_0}{N}\right) \]
where \(t = 1 \, \text{hour}\).
2. **Calculate \(\lambda\)**:
\[ \lambda = \frac{2.303}{1} \log\left(\frac{6.022 \times 10^{24}}{5.721 \times 10^{24}}\right) = 2.303 \log(1.0527) \approx 2.303 \times 0.0227 \approx 0.0522 \, \text{hour}^{-1} \]

### Step 6: Calculate the half-life (\(t_{1/2}\)).
1. **Half-life Formula**:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
2. **Calculate \(t_{1/2}\)**:
\[ t_{1/2} = \frac{0.693}{0.0522} \approx 13.27 \, \text{hours} \]

### Final Answer:
The half-life of the radioisotope is approximately **13.27 hours**.