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NEET generalGeneralMedium

Question

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 kg2.76 \times 10^{-26} \text{ kg}, Boltzmann's constant kB=1.38×10−23 JK−1k_B = 1.38 \times 10^{-23} \text{ JK}^{-1})

A

5.016×104 K5.016 \times 10^4 \text{ K}

B

8.360×104 K8.360 \times 10^4 \text{ K}

C

2.508×104 K2.508 \times 10^4 \text{ K}

D

1.254×104 K1.254 \times 10^4 \text{ K}

Step-by-Step Solution

The escape velocity Vescape=11200 m/sV_{escape} = 11200 \text{ m/s}. The rms speed is given by 3kBTm=11200 m/s\sqrt{\frac{3k_B T}{m}} = 11200 \text{ m/s}. Solving for T gives T=8.360×104 KT = 8.360 \times 10^4 \text{ K}.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generaltemperatureoxygenmoleculesbecomesufficient

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