Slit width for minimum spot size in electron slit diffraction if involving uncertainity principle

Question

I don't believe the following is an accurate description of the physical but a homework problem to help understanding.

A beam of electron of energy 0.025 eV moving along x-direction, passes through a slit of variable width w placed along y-axis. Estimate the value of the width of the slit for which the spot size on a screen kept at a distance of 0.5 m from slit would be minimum.

I have the following idea:

$$\theta \sim \frac{\lambda}{d}$$, where $\lambda$ is the wavelength, d is slit width, and $\theta$ represents diffraction spot size as an angle measure. For any $L$, the distance to the screen, this quantity goes minimum for $d$ tending to $\infty$

I have an intuition that the uncertainty principle shall be used to get a upper bound on $d$ as effects of Heisenberg's uncertainty exceed effects of diffraction.

That way, I would have $$v_y = \Delta v_y = \frac{\hbar}{m_e \Delta y} = \frac{\hbar}{m_e d}$$

Taking $$v_x = \sqrt{\frac{2 K}{m_e}}$$, we have the spot size in angles as

$$\frac{v_y}{v_x} = \frac{\hbar}{d \sqrt{2 K m_e}} = \frac{\lambda}{d}$$

That leaves me nowhere only when it starts seeming I have solved it.

Please help.

Answer 1

If the slit width is large compared to the electron wavelength then the spot size will be the same as the slit width (assuming the electron beam doesn't diverge):

$$ \text{spot size} = w $$

In the limit of small slit width you get the equation you cite for the (half) angular divergence, and the spot width will be (assuming $\sin\theta \approx \theta$):

$$ \text{spot size} = 2L\frac{\lambda}{w} $$

where $L$ is the distance to the screen (0.5m). The smallest spot will be when these coincide i.e.

$$ w = 2L\frac{\lambda}{w} $$

or:

$$ w = \sqrt{2L\lambda} $$

which I get to be about 88 microns. NB this is a very rough calculation as the spot doesn't have a clearly defined width and even with large slits there will be diffraction at the slit edges.