- #1
limper
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Homework Statement
Some particles pass through a single slit of width W = 0.17 mm. After the particles pass through the slit they spread out over a range of angles. The de Broglie wavelength of each particle is λ = 561 nm. Use the Heisenberg uncertainty principle to determine the minimum range of angles.
θ = (the diagram shows theta as the entire area between the first two minima, not just from the center to one side)
Homework Equations
It seems like this would just be normal diffraction with sin()=λ/d.
The problem asks for it to be done via the uncertainty principle so:
ΔyΔpy>=h/4π
Δpy=psin()
p=h/λ
The Attempt at a Solution
sin()=λ/d yields (561*10^-9)/(.17*10^-3)=.0033 rad, or .0066 rad for the entire range.
The other method gives the same result if you use just h, without the 1/4π, for the uncertainty equation. Is there a reason for that? I know h/4π is the lower bound, is there something inherent to the problem that limits it to h?
otherwise:
ΔyΔpy>=h/4π
Δpy>=h/(4πΔy)
Δpy>=3.1*10^-31
p=h/λ
p=(6.626*10^-34)/(561*10^-9)
p=1.2*10^-27
sin()=Δpy/p
sin()=(3.1*10^-31)/(1.2*10^-27)
=2.6*10^-4 rad or 5.2*10^-4 rad for the entire range
This seems to make sense to me, but the homework software is rejecting both answers. I've tried it with answers past the significant digits and no dice, so I'm wondering if I've misunderstood something or made a silly mistake somewhere.