Electron Single Slit Diffraction

[hy23]

Homework Statement

Electrons pass through a 10^-6m wide slit with a speed of 1.5 x 10⁶m/s. How wide is the electron diffraction pattern on a detector 1 m behind the slit.

Homework Equations

$$\lambda$$=h/mv (matter waves)
asin$$\theta$$=m$$\lambda$$ (path length difference for constructive interference)
y=Ltan$$\theta$$ (location of bright fringes)

The Attempt at a Solution

First I found the wavelength of the electrons using De Broglie's equation. This I'm pretty sure is correct. What I'm not sure about is how electrons diffract in a single slit. i don't know if the formulas for the single slit diffraction of light waves will work for electrons.

 

[vela]

Diffraction is a wave phenomenon. It doesn't matter what kind of wave, so the equations you had for diffraction with light will work. The equation you have above, however, is for double-slit interference.

 

[hy23]

what??/ really? my book says that's the equation for single slit...in double slit it is different in that they use the small angle approximation

what is the formula for single slit then

and the other thing i was wondering about is that won't there be a LOT of bright bringes and thus the diffraction pattern can extend really far so how would we know which m value to use?

 

[vela]

The problem is probably asking for the width of the central peak.

http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction

 

[rwisz]

I also don't know how to find the angle theta for the electrons as they don't have a frequency or wavelength.

I can understand your confusion about applying the formulas for single slit diffraction of light waves to electrons. However, it is important to note that electrons also exhibit wave-like behavior, known as matter waves. Therefore, the same principles of diffraction and interference that apply to light waves can also be applied to electrons.

To find the width of the diffraction pattern on the detector, we can use the formula for the location of bright fringes, y = Ltanθ. In this case, the distance L is 1 m and we need to find the value of θ. To do so, we can use the formula asinθ = mλ, where m is the order of the bright fringe and λ is the wavelength of the electrons.

We have already calculated the value of λ using De Broglie's equation, which is correct. Now, we need to determine the value of θ for the first order bright fringe, which is m=1. Therefore, we can rearrange the equation asinθ = mλ to solve for θ.

θ = asin(mλ) = asin(1*λ) = asin(1*6.63x10^-34/1.5x10^6) = 2.82x10^-8 radians.

Now, we can plug in this value of θ into the formula for the location of bright fringes, y = Ltanθ, to find the width of the diffraction pattern on the detector.

y = (1m) * tan(2.82x10^-8 radians) = 2.82x10^-8 m

Therefore, the width of the electron diffraction pattern on the detector will be approximately 2.82x10^-8 m.