Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-

Solution

Physics sir · 3 steps

Same temperature means same average kinetic energy, so the heavier proton carries more momentum.

1. Thermal momentum. In an ideal gas at temperature T, the average kinetic energy is 3/2kT for every particle. Since p²/2m = 3/2kT, the typical momentum is p = √3mkT.
2. Uncertainty in position. The spread in momentum is of the order of the momentum itself, so Δ p ∼ p. The uncertainty principle then gives Δ x ∼ h/(Δ p) ∝ 1/√m, which is the same as the de Broglie wavelength λ = h/√3mkT.

   (Δ xₑ)/(Δ xₚ) = √(3mₚ kT)/√(3mₑ kT) = √(mₚ/mₑ)

3. So. The lighter electron has less momentum, so its position is more uncertain. The ratio is proportional to √(mₚ/mₑ), which is option (c).

The idea to keep. For thermal particles, p = √3mkT, so both λ and Δ x go as 1/√mT. Same temperature means lighter particle, larger uncertainty.

Answer from the NTA final answer key. Solution by PrepWiser.