An electron moving in the positive $x$ -direction passes through a slit of width $\Delta y=85 \mathrm{nm} .$ What is the minimum uncertainty in the electron's velocity in the $y$ -direction?
Added by Josefina H.
Step 1
Step 1: Recall the formula for uncertainty in velocity in the y-direction: $\Delta v_y = \frac{\hbar}{2m} \Delta y$, where $\hbar$ is the reduced Planck constant, $m$ is the mass of the electron, and $\Delta y$ is the width of the slit. Show more…
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