If escape velocity from the earth is 11.1 km/s and the mass of one molecule of oxygen is `5.34 xx 10^(-26)` kg, the temperature at which the oxygen molecule will escape from earth, is [Boltzmann constant `k=1.38xx10^(-23)` J/K]

Options

A. `1.6xx10^5` K
B. `1.6xx10^3` K
C. `1.6xx10^2` K
D. None of these

Answer

To find the temperature at which an oxygen molecule can escape from Earth, we can use the relationship between kinetic energy and temperature. The escape velocity (v_e) is given, and we can relate it to the kinetic energy of the molecule and the temperature using the following steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Escape velocity, \( v_e = 11.1 \, \text{km/s} = 11.1 \times 10^3 \, \text{m/s} \)
- Mass of one molecule of oxygen, \( m = 5.34 \times 10^{-26} \, \text{kg} \)
- Boltzmann constant, \( k = 1.38 \times 10^{-23} \, \text{J/K} \)

2. **Calculate the kinetic energy required to escape:**
The kinetic energy (KE) needed for an object to reach escape velocity is given by:
\[ KE = \frac{1}{2} m v_e^2 \]

3. **Substituting the values into the kinetic energy formula:**
\[ KE = \frac{1}{2} \times (5.34 \times 10^{-26} \, \text{kg}) \times (11.1 \times 10^3 \, \text{m/s})^2 \]

4. **Calculating \( v_e^2 \):**
\[ v_e^2 = (11.1 \times 10^3)^2 = 1.2321 \times 10^8 \, \text{m}^2/\text{s}^2 \]

5. **Now calculate the kinetic energy:**
\[ KE = \frac{1}{2} \times (5.34 \times 10^{-26}) \times (1.2321 \times 10^8) \]
\[ KE = \frac{1}{2} \times (6.573 \times 10^{-18}) = 3.2865 \times 10^{-18} \, \text{J} \]

6. **Relate kinetic energy to temperature:**
The kinetic energy can also be expressed in terms of temperature using the formula:
\[ KE = \frac{3}{2} k T \]

7. **Setting the two expressions for kinetic energy equal:**
\[ 3.2865 \times 10^{-18} = \frac{3}{2} (1.38 \times 10^{-23}) T \]

8. **Solving for temperature \( T \):**
\[ T = \frac{2 \times 3.2865 \times 10^{-18}}{3 \times 1.38 \times 10^{-23}} \]
\[ T = \frac{6.573 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 1.587 \times 10^5 \, \text{K} \]

9. **Final result:**
\[ T \approx 1.56 \times 10^5 \, \text{K} \]

### Conclusion:
The temperature at which the oxygen molecule will escape from Earth is approximately \( 1.56 \times 10^5 \, \text{K} \).