All Problems

Wave Properties of particles. Schrodinger Equation.

Problem 6.75

A parallel stream of hydrogen atoms with velocity vv =600 m/s=600 \mathrm{~m} / \mathrm{s} falls normally on a diaphragm with a narrow slit behind which a screen is placed at a distance l=1.0 m.l=1.0 \mathrm{~m} . Using the uncertainty principle, evaluate the width of the slit δ\delta at which the width of its image on the screen is minimum.

Reveal Answer
6.75. The width of the image is Δ≈δ+Δ′≈δ+ℏl/pδ\Delta \approx \delta+\Delta^{\prime} \approx \delta+\hbar l / p \delta, where Δ′\Delta^{\prime} is an additional widening associated with the uncertainty of the momentum Δpy\Delta p_{y} (when the hydrogen atoms pass through the slit), pp is the momentum of the incident hydrogen atoms. The function Δ(δ)\Delta(\delta) has the minimum when δ≈ℏl/mv=0.01 mm\delta \approx \sqrt{\hbar l / m v}=0.01 \mathrm{~mm}.
Prev
Next