Uncertainty -- particle through single slit

[limper]

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.

 

[drvrm]

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,

i do not know /can not guess about the software you are using but the angular spread comes out to be wavelength /slit width using uncertainty relation.
i think the uncertainty be taken of the order of h rather than minimum of the product h/4pi. as the wave packet is not being defined for exact measurement of the product.