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.
Questions Using Planck's Constant
An electron microscope operates with a beam of electrons, each of which has an energy of 20 KeV. Use the uncertainty principle in the form delta(x)delta(p) (greater or equal to) h/2 to find the smallest size that such a device could resolve. Planck's constant is 1.0552 × 10^-34 J · s. Answer in units of pm.
What must the energy of each neutron in a beam of neutrons be in order to resolve the same size object?
Answer in units of eV.
A beam of neutrons with a kinetic energy of 0.0002 eV falls on a slit of width 0.0001 m.
The Planck's constant is 6.63 × 10 ^-34 J · S.
What will be the angular spread of the beam after it passes through the slit?
Answer in units of radian.View Full Posting Details