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