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The temperature at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to [Boltzmann constant k_B=1.38 × 10⁻²³ J/K, Avogadro Number N_A=6.02 × 10²⁶ /kg, Radius of Earth = 6.4 × 10⁶ m, g on Earth = 10 m s⁻²]

Asked in JEE Main 8th April 2nd Shift 2019 · Escape speed and Brownian motion

Answer: (1) 10⁴ K

Step-by-step solution

Escape speed from the surface: v_esc=√2gR=√2×10×6.4×10⁶.

v_esc²=1.28×10⁸ m²/s².

The given N_A=6.02×10²⁶ per kilogram-mole, with M=2 kg for H₂, gives m=2/(6.02×10²⁶)=3.32×10⁻²⁷ kg.

Set (3k_BT)/m=v_esc², so T=(mv_esc²)/(3k_B).

T=(3.32×10⁻²⁷×1.28×10⁸)/(3×1.38×10⁻²³)=(4.25×10⁻¹⁹)/(4.14×10⁻²³)≈1.0×10⁴ K.

Why the other options are wrong

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