Question

Exercise 7:

Let’s imagine a SINGLE SLIT experiment with electrons of energy 10.0eV (probably too little to get it through a REAL experiment, but hey, we’re imagining.) Suppose the slit width is 20.0nm. See figures 24-21 and 24-22 to remember single slit diffraction. Fig 24-22 is in fact what I am imagining for this problem.A) What is the angle to the first diffraction minimum on either side of the center? You must find the wavelength of these electrons, but it’s the same calculation you did in Exercise 11 of HW7, by first getting momentum and then wavelength. I get about 1 degree.

B) Adapting Figure 24-22, we see an incident beam of electrons with horizontal momentum yp(they chose x as vertical for this problem) hitting the slit, but afterward the electrons must have a spread of vertical (x) momentum components in order to hit the screen and make the spread shown. How does the spread in vertical momentum components change as the slit width is made narrower? (increase, decrease, or remain more or less the same?)

C) According to the uncertainty principle, when we try to force a quantum particle into a smaller region of space, the range of possible momentum values increases. How does that concept apply to this single-slit diffraction case?

D) Finally, let’s assume that “Δx” is the slit width and “Δpx” is the x-component of the momentum of these electrons at the angle of the first minimum. For simplicity, take the momentum from part A and find the vertical component using the angle. Show that the Heisenberg uncertainty principle is satisfied. [Note: In classes on quantum theory, an exact statistical definition of uncertainty is applied. Usually it is the “variance,” for those of you who know statistics. In the case here, we simply know that MOST electrons will fall in the center of the diffraction pattern, so we may take the width of the region between the first minima as representing the spread in (vertical) momentum. Remember (from class) there is no restriction on the vertical position and horizontal momentum.

Answer 1

A)

For first minimum

a=slit width

or

B)

If slit width then from ,   then vertical component    increases.

C)

uncertainity principle says

If uncertaining in x is decreased then uncertainty in increases k this concept satisfy the single slit diffraction, Here   and

D)

and

Then

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