1. 2.508×10⁴ K 2. 5.016×10⁴ K 3. 8.360×10⁴ K 4. 1.254×10⁴ K

Answer: 3. 8.360×10⁴ K The molecules escape only if their rms speed equals the escape speed of the Earth, v_e = 11.2\ mathrmkm s⁻¹ = 1.12×10⁴\ mathrmm s⁻¹ . v_rms=√(3k_BT/m) so T=m v_e²/3k_B . T=frac2.76×10⁻²⁶×(1.12×10⁴)²3×1.38×10⁻²³≈ 8.36×10⁴\ K The correct option is 8.360×10⁴ K.

Topic: Kinetic interpretation of temperature, RMS speed· Kinetic Theory of Gases

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) = 2.76×10^-26 kgBoltzmann's constant k_B = 1.38×10^-23 J K^-1)

1. $2.508\times10^4$ K

2. $5.016\times10^4$ K

3. $8.360\times10^4$ K

4. $1.254\times10^4$ K

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 3. $8.360\times10^4$ K

The molecules escape only if their rms speed equals the escape speed of the Earth, $v_e = 11.2\ \mathrm{km\,s^{-1}} = 1.12\times10^4\ \mathrm{m\,s^{-1}}$.

$v_{rms}=\sqrt{\dfrac{3k_BT}{m}}$ so $T=\dfrac{m v_e^2}{3k_B}$.

$$T=\frac{2.76\times10^{-26}\times(1.12\times10^{4})^2}{3\times1.38\times10^{-23}}\approx 8.36\times10^{4}\ \mathrm{K}$$

The correct option is $8.360\times10^4$ K.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

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) = 2.76×10^-26 kgBoltzmann's constant k_B = 1.38×10^-23 J K^-1)

1 answer

Your answer