A beam of electrons traveling with speed 7 × 10 ^7 m/s passes through a slit of width 1 × 10 ^(-5) m. Because of the uncertainty in the lateral position of the beam, there will be an uncertainty in the transverse momentum as well.
Estimate this uncertainty, and use it to calculate the spread of the image of the electron
beam on a photographic plate placed perpendicular to the beam at a distance of 2 m
beyond the slit.
The uncertainty relation between momentum and positition is:
Delta x Delta P >= h-bar/2 (1)
Here Delta P is the uncertainty of a component of the momentum, say, the x-component, and Delta x is then the uncertainty of the position in the x-direction. The constant h-bar is the so-called reduced Planck's constant, it is just the ordinary Plancks constant divided by 2 pi. So, h-bar = 1.05x10^(-34) J s
In this problem we can take Delta x = 1 x 10 ^(-5) m. The electrons, when they are inside the slit, can have any position compatible with them being in the slit. So, their lateral component of the position has an uncertainty of
Delta x = 1 x 10 ^(-5) m. But we may object to this and say that we can imagine ...
A detailed solution is given.