The temperature ,a t which the root mean square velcity of hydrogen molecules equals their escape velocity from the earth, is closed to:
[Boltzmann Constant `k_(B)=1.38xx10^(-23)j//K` Avogadro Number `N_(A)=6.02xx10^(26)//Kg`
Radius of Earth :`6.4xx10^(6)m`
Gravitational acceleration
On earth =`10ms^(-2)]`

Options

A. `800 K`
B. `3xx10^(5)K`
C. `10^(4)K`
D. `650K`

Answer

To solve the problem, we need to find the temperature at which the root mean square (RMS) velocity of hydrogen molecules equals their escape velocity from the Earth.

### Step 1: Write the formula for escape velocity
The escape velocity \( v_e \) from the surface of the Earth can be calculated using the formula:
\[ v_e = \sqrt{2gr} \]
where:
- \( g \) is the gravitational acceleration (10 m/s²),
- \( r \) is the radius of the Earth (6.4 × 10⁶ m).

### Step 2: Calculate escape velocity
Substituting the values into the escape velocity formula:
\[ v_e = \sqrt{2 \times 10 \, \text{m/s}^2 \times 6.4 \times 10^6 \, \text{m}} = \sqrt{128000000} \approx 11314.0 \, \text{m/s} \]

### Step 3: Write the formula for root mean square velocity
The root mean square velocity \( v_{rms} \) of hydrogen molecules can be expressed as:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- \( k \) is the Boltzmann constant (1.38 × 10⁻²³ J/K),
- \( T \) is the temperature in Kelvin,
- \( m \) is the mass of a hydrogen molecule.

### Step 4: Convert molar mass to mass of a single molecule
The molar mass of hydrogen is approximately 2 g/mol. To convert this to kilograms:
\[ m = \frac{2 \times 10^{-3} \, \text{kg}}{N_A} \quad \text{(where \( N_A \) is Avogadro's number, \( 6.02 \times 10^{23} \, \text{mol}^{-1} \))} \]
Calculating \( m \):
\[ m = \frac{2 \times 10^{-3}}{6.02 \times 10^{23}} \approx 3.32 \times 10^{-27} \, \text{kg} \]

### Step 5: Set RMS velocity equal to escape velocity
Since we need to find the temperature where \( v_{rms} = v_e \):
\[ \sqrt{\frac{3kT}{m}} = \sqrt{2gr} \]

### Step 6: Square both sides
Squaring both sides gives:
\[ \frac{3kT}{m} = 2gr \]

### Step 7: Solve for temperature \( T \)
Rearranging the equation to solve for \( T \):
\[ T = \frac{2grm}{3k} \]

### Step 8: Substitute known values
Substituting \( g = 10 \, \text{m/s}^2 \), \( r = 6.4 \times 10^6 \, \text{m} \), \( m = 3.32 \times 10^{-27} \, \text{kg} \), and \( k = 1.38 \times 10^{-23} \, \text{J/K} \):
\[ T = \frac{2 \times 10 \times 6.4 \times 10^6 \times 3.32 \times 10^{-27}}{3 \times 1.38 \times 10^{-23}} \]

### Step 9: Calculate the temperature
Calculating the numerator:
\[ 2 \times 10 \times 6.4 \times 10^6 \times 3.32 \times 10^{-27} \approx 4.23 \times 10^{-19} \]
Calculating the denominator:
\[ 3 \times 1.38 \times 10^{-23} \approx 4.14 \times 10^{-23} \]
Now substituting:
\[ T \approx \frac{4.23 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 1.02 \times 10^4 \, \text{K} \]

### Conclusion
The temperature at which the root mean square velocity of hydrogen molecules equals their escape velocity from the Earth is approximately:
\[ T \approx 1.01 \times 10^4 \, \text{K} \]