At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?
(Given:Mass of oxygen molecule (m) = `2.76xx10^(-26)` kg, Boltzmann's constant `k_B-1.38xx10^(-23)JK^(-1)`

Options

A. `2.508xx10^(4)K`
B. `1.254xx10^(4)K`
C. `5.016xx10^(4)K`
D. `8.360xx10^(4)K`

Answer

To solve the problem of determining the temperature at which the root mean square (RMS) speed of oxygen molecules is sufficient for escaping from the Earth's atmosphere, we will follow these steps:

### Step 1: Understand the formulas involved
1. **Escape Velocity (v_escape)**: The formula for escape velocity from the Earth's surface is given by:
\[ v_{\text{escape}} = \sqrt{\frac{2GM_E}{R_E}} \]
where:
- \( G \) = gravitational constant \( \approx 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- \( M_E \) = mass of the Earth \( \approx 5.97 \times 10^{24} \, \text{kg} \)
- \( R_E \) = radius of the Earth \( \approx 6.37 \times 10^6 \, \text{m} \)

2. **RMS Speed (v_rms)**: The formula for the RMS speed of gas molecules is given by:
\[ v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} \]
where:
- \( k_B \) = Boltzmann's constant \( = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) = temperature in Kelvin
- \( m \) = mass of the oxygen molecule \( = 2.76 \times 10^{-26} \, \text{kg} \)

### Step 2: Set the equations equal
To find the temperature at which the RMS speed equals the escape velocity, we set the two equations equal:
\[ \sqrt{\frac{2GM_E}{R_E}} = \sqrt{\frac{3k_B T}{m}} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square roots gives:
\[ \frac{2GM_E}{R_E} = \frac{3k_B T}{m} \]

### Step 4: Solve for T
Rearranging the equation to solve for \( T \):
\[ T = \frac{2GM_E m}{3k_B R_E} \]

### Step 5: Substitute the known values
Now we will substitute the known values into the equation:
- \( G = 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
- \( M_E = 5.97 \times 10^{24} \, \text{kg} \)
- \( m = 2.76 \times 10^{-26} \, \text{kg} \)
- \( k_B = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( R_E = 6.37 \times 10^6 \, \text{m} \)

Substituting these values:
\[ T = \frac{2 \times (6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (2.76 \times 10^{-26})}{3 \times (1.38 \times 10^{-23}) \times (6.37 \times 10^6)} \]

### Step 6: Calculate the temperature
Calculating the above expression step-by-step:
1. Calculate the numerator:
- \( 2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 2.76 \times 10^{-26} \)
- This gives approximately \( 2.77 \times 10^{-12} \)

2. Calculate the denominator:
- \( 3 \times 1.38 \times 10^{-23} \times 6.37 \times 10^6 \)
- This gives approximately \( 2.63 \times 10^{-16} \)

3. Now divide the numerator by the denominator:
\[ T \approx \frac{2.77 \times 10^{-12}}{2.63 \times 10^{-16}} \approx 105.3 \times 10^4 \, \text{K} \approx 8.36 \times 10^4 \, \text{K} \]

### Final Answer
The temperature at which the RMS speed of oxygen molecules becomes sufficient for escaping from the Earth's atmosphere is approximately:
\[ T \approx 8.36 \times 10^4 \, \text{K} \]