A krypton atom is released from the surface of the earth to travel upwards at 450 K, moving with its most probable speed. Assuming that it undergoes no collisions with other molecules, how far (in met

Question

A krypton atom is released from the surface of the earth to travel upwards at 450 K, moving with its most probable speed. Assuming that it undergoes no collisions with other molecules, how far (in meter) will it travel before coming to rest? (Atomic mass of Kr = 84, g = 10 ms⁻², R = 8.4 J/K-mol)

TestHub · Accepted Answer

The most probable speed is vₚ = sqrt(2RT/M) = sqrt(2 × 8.4 × 450/0.084) = sqrt(90000) = 300 m/s. Rising against gravity, the atom stops when (1/2)mv² = mgh, so h = vₚ²/(2g) = 90000/(2 × 10) = 4500 m. (Trap: using the rms speed instead would give 3RT/(2Mg) = 6750 m; the stem specifies the most probable speed.) Hence the distance is 4500 m.

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