An electron is sent through a slit whose width is $\Delta$x = 0.040 cm.The uncertainty in the x component of the momentum of this electron is approximately A)1.3 $\times$ 10^-24 N · s B)2.6 $\times$ 10^-24 N · s C)1.3 $\times$ 10^-23 N · s D)2.6 $\times$ 10^-23 N · s E)3.9 $\times$ 10^-25 N · s

Question

Quizplus · Accepted Answer

The uncertainty in the momentum ($\Delta p_x$) can be estimated using Heisenberg's uncertainty principle, which states $\Delta x \Delta p_x \approx h$, where $h$ is Planck's constant ($6.626 \times 10^{-34} m^2 kg / s$). Given $\Delta x = 0.040 cm = 0.0004 m$, solving for $\Delta p_x$ gives a value close to $1.3 \times 10^{-24} N \cdot s$.

An Electron Is Sent Through a Slit Whose Width Is Δ\DeltaΔ

An Electron Is Sent Through a Slit Whose Width Is $\Delta$