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If escape velocity from the earth is 11....

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]

A

`1.6xx10^5` K

B

`1.6xx10^3` K

C

`1.6xx10^2` K

D

None of these

Text Solution

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The correct Answer is:
To find the temperature at which an oxygen molecule will escape from Earth, we can use the relationship between kinetic energy and temperature. The escape velocity (V_e) is given as 11.1 km/s, and the mass of one molecule of oxygen (m) is given as \(5.34 \times 10^{-26}\) kg. The Boltzmann constant (k) is \(1.38 \times 10^{-23}\) J/K. ### Step-by-Step Solution: 1. **Convert Escape Velocity to Meters per Second**: \[ V_e = 11.1 \text{ km/s} = 11.1 \times 10^3 \text{ m/s} \] 2. **Write the Kinetic Energy Expression**: The kinetic energy (KE) of a molecule is given by: \[ KE = \frac{1}{2} m V_e^2 \] 3. **Set Kinetic Energy Equal to Escape Energy**: The kinetic energy required for the molecule to escape can be equated to the thermal energy: \[ KE = \frac{3}{2} k T \] 4. **Equate the Two Expressions**: \[ \frac{1}{2} m V_e^2 = \frac{3}{2} k T \] 5. **Solve for Temperature (T)**: Rearranging the equation gives: \[ T = \frac{m V_e^2}{3k} \] 6. **Substitute the Values**: Substitute \(m = 5.34 \times 10^{-26} \text{ kg}\), \(V_e = 11.1 \times 10^3 \text{ m/s}\), and \(k = 1.38 \times 10^{-23} \text{ J/K}\): \[ T = \frac{(5.34 \times 10^{-26}) (11.1 \times 10^3)^2}{3 \times (1.38 \times 10^{-23})} \] 7. **Calculate \(V_e^2\)**: \[ V_e^2 = (11.1 \times 10^3)^2 = 1.2321 \times 10^8 \text{ m}^2/\text{s}^2 \] 8. **Calculate the Numerator**: \[ \text{Numerator} = 5.34 \times 10^{-26} \times 1.2321 \times 10^8 = 6.578 \times 10^{-18} \] 9. **Calculate the Denominator**: \[ \text{Denominator} = 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \] 10. **Calculate Temperature (T)**: \[ T = \frac{6.578 \times 10^{-18}}{4.14 \times 10^{-23}} \approx 1.59 \times 10^5 \text{ K} \] ### Final Result: The temperature at which the oxygen molecule will escape from Earth is approximately: \[ T \approx 1.59 \times 10^5 \text{ K} \]
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