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NEET PHYSICSMedium

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 kgm = 2.76 \times 10^{-26} \text{ kg}, Boltzmann's constant kB=1.38×10−23 J K−1k_B = 1.38 \times 10^{-23} \text{ J K}^{-1})

A

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

B

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

C

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

D

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

Step-by-Step Solution

To escape Earth's atmosphere, the molecule's root mean square (rms) speed must equal the Earth's escape velocity.

  1. Formulas: RMS Speed (vrmsv_{rms}) = 3kBTm\sqrt{\frac{3 k_B T}{m}} Escape Velocity (vev_{e}) for Earth ≈11.2 km/s=11200 m/s\approx 11.2 \text{ km/s} = 11200 \text{ m/s}

  2. Set up the equation: 3kBTm=ve\sqrt{\frac{3 k_B T}{m}} = v_{e} Squaring both sides: 3kBTm=ve2\frac{3 k_B T}{m} = v_{e}^2 T=mve23kBT = \frac{m v_{e}^2}{3 k_B}

  3. Substitute values: m=2.76×10−26 kgm = 2.76 \times 10^{-26} \text{ kg} ve=11200 m/sv_{e} = 11200 \text{ m/s}

  • kB=1.38×10−23 J K−1k_B = 1.38 \times 10^{-23} \text{ J K}^{-1}

T=(2.76×10−26)×(11200)23×(1.38×10−23)T = \frac{(2.76 \times 10^{-26}) \times (11200)^2}{3 \times (1.38 \times 10^{-23})} T=2.76×1.2544×108×10−264.14×10−23T = \frac{2.76 \times 1.2544 \times 10^8 \times 10^{-26}}{4.14 \times 10^{-23}} T=3.462×10−184.14×10−23T = \frac{3.462 \times 10^{-18}}{4.14 \times 10^{-23}} T≈0.8362×105 KT \approx 0.8362 \times 10^5 \text{ K} T≈8.362×104 KT \approx 8.362 \times 10^4 \text{ K}

This matches Option 2.

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