At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?(Given: Mass of oxygen mole - Physics MCQ NEET (UG) 2018 2026 | ExamTest.live

hardMCQNEET (UG) 20182026Physics

1 mark

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 1 0^{- 26} kg, Boltzmann's constant k_{B} = 1.38 \times 1 0^{- 23} JK^{- 1}, Escape velocity v_{e} = 11200 m/s)

Solution & Step-by-step Explanation

The root mean square (rms) speed is given by: v_{r m s} = \frac{3 k _{B} T}{m} To escape the atmosphere, v_{r m s} must equal the escape velocity v_{e} : v_{e} = \frac{3 k _{B} T}{m} v_{e}^{2} = \frac{3 k _{B} T}{m} T = \frac{m v _{e}^{2}}{3 k _{B}} Substitute the values: T = \frac{( 2.76 \times 1 0 ^{- 26} ) \times ( 11200 ) ^{2}}{3 \times ( 1.38 \times 1 0 ^{- 23} )} T = \frac{2.76 \times 1 0 ^{- 26} \times 1.2544 \times 1 0 ^{8}}{4.14 \times 1 0 ^{- 23}} T \approx 8.36 \times 1 0^{4} K

Try it yourself before checking the explanation above.

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 1 0^{- 26} kg, Boltzmann's constant k_{B} = 1.38 \times 1 0^{- 23} JK^{- 1}, Escape velocity v_{e} = 11200 m/s)

5.016 \times 1 0^{4} K

8.360 \times 1 0^{4} K

2.508 \times 1 0^{4} K

1.254 \times 1 0^{4} K

Practice with timed mock tests and track your performance across Physics.