Question on Heisenberg uncertainty principle

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SUMMARY

The discussion focuses on applying the Heisenberg uncertainty principle to determine the minimum range of angles for particles passing through a single slit of width 0.2 mm, with a de Broglie wavelength of 633 nm. The relevant equation is sin θ = λ/W, where θ represents the angle of diffraction. Participants clarify that the uncertainty in position can be approximated using half the slit width, leading to an estimation of the angle range from -π/2 to +π/2, although the distribution becomes negligible beyond certain angles. The conversation emphasizes using the uncertainty principle as an estimation tool rather than relying solely on the single slit diffraction equation.

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Homework Statement


Particles pass through a single slit of width 0.2 mm in a diffraction setup. The de Broglie wavelength of each particle is 633 nm. After the particles pass through the slit, they spread out over a range of angles. Use the Heisenberg uncertainty principle to determine the minimum range of angles.


Homework Equations


Heisenberg uncertainty principle
sin \theta= \lambda/W


The Attempt at a Solution


I'm not sure exactly what a minimum range of angles mean. Is it all the possible angles through which the electrons can follow?
I believe I'm stuck on how to use the uncertainty principle to find an angle. I know that the equation for diffraction through a single slit might be useful.
 
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Well, the wording of these sort of statements can be tricky. The particle distribution will actually extend through the full -π/2 to +π/2 range. However, the distribution becomes negligibly small beyond some angle.

Strictly speaking, "uncertainty" in physics usually means an rms deviation from a central value (in this case, 0 radians). Since, in this case, we are going after a ballpark figure, it's probably okay to use half the slit width as the initial position uncertainty.

The idea is to use the Heisenberg relation as an estimate, rather than the single slit diffraction equation. Of course, you are welcome to compare the two results afterward--in which case one might use the angle where the diffraction equation gives 1/2 of the peak value.
 
I see, so the distribution includes the negative angles.
Thank you for a clear explanation!
 

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