Q. 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) = $2.76 \times 10^{-26}$ kg Boltzmann's constant $k_B = 1.38 \times 10^{-23}$ J $K^{-1}$)
A) $5.016 \times 10^4$ K
B) $8.360 \times 10^4$ K
C) $2.508 \times 10^4$ K
D) $1.254 \times 10^4$ K
Explanation
[Solution] $\sqrt{3k_BT/m} = v_e = 11.2$ km/s gives $T = \frac{mv_e^2}{3k_B} = \frac{2.76 \times 10^{-26} \times (11.2 \times 10^3)^2}{3 \times 1.38 \times 10^{-23}} \approx 8.36 \times 10^4$ K. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.