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Given an electron with mass $9.1\times10^{-31}kg$ is moving along $x$-axis with velocity $10^{6} ms^{-1}$ and passes through a screen with slit $\Delta y=10^{-9}$. Is it possible to find the uncertainty of angle between the final vector and the $x$-axis?

My attempt:

  1. Finding the uncertainty of momentum $\Delta p$ by using the Heisenberg's uncertainty principle.
  2. Solving the uncertainty of angle with the momentum along x-direction $p_x$, by using $$tan(\Delta \theta)=\frac{\Delta p_y}{p_x}$$

I doubt step(2) because we are looking for uncertainty but not the angle itself. I have no idea about the uncertainty.

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I think you are missing something.

According to Heisenberg's uncertainty principle,

$$\Delta y\Delta p_y\sim\hbar$$

If $\Delta v_y$ is the uncertainty in velocity along the $y$- direction, then

$$ \begin{align} m\Delta y \Delta v_y&\sim\hbar \\ m\Delta yv_x\Delta \theta&\sim\hbar\\ \Delta\theta&\sim\frac{\hbar}{mv_x\Delta y} \end{align} $$

where we have taken $\Delta \theta$ be the uncertainty in the angle between the $x$- and $y$- component of velocities.

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