A small particle of mass moves along the x-axis

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Prince Budha · Accepted Answer

Solution: Here, given mass of small particle(m) = 10-6 g = 10-6 × 10-3 kg Speed of particle(v) = 10-6 ms-1 Uncertainty in the x-coordinate of the particle(Δx) = ? Uncertainty in velocity(Δx)= 10-6 ms-1 We know that, from Heisenberg's uncertainty principle, Δx Δp ≥ h or, Δx Δp = h/2π a. Here, Δp = mparticle Δv (∴ Δx = frac{h}{2pi Delta p}) (= frac{h}{2pi m Delta v}) (= frac{6.626 times 10^{-34}}{2pi times 10^{-9} times 10^{-6}}) (= 1.05 times 10^{-19} m) (therefore Delta x = 1.05 times 10^{-19} m.) b. Mass of electron me = 9.1 × 10-31 kg Then, uncertainty in x-coordinate or position of electron (Delta x = frac{h}{2pi Delta p}) (= frac{h}{2pi mDelta v}) (= frac{6.626 times 10^{-34}}{2pi times 9.1 times 10^{-31} times 10^{-6}}) = 115.768 (therefore Delta x = 115.768 approx 116 m.)

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