How Does Electron Wavelength Affect Two-Slit Experiment Outcomes?

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Homework Statement

Consider a two-slit experiment using electrons. The slits are assumed to be very narrow compared to the wave-length of the electrons. Beyond the slits is a bank of electron detectors. At the center detector, directly in the path the beam would follow if unobstructed, 100 electrons per second are detected. Suppose that as the detector angle varies, the number per unit time of electron’s arriving varies from a maximum of 100/s to a minimum of 0. Suppose the electrons have K = 1.0 eV of kinetic energy and the narrow slits are separated by 0.020 μm.
a/ At what angle is the detector X located where the minimum is reached ?
b/ How many electrons would be detected per second at the center detector if one of the slits were blocked?
c/ How many electrons would be detected per second at the center detector and at detector X if one of the slits were narrowed to 36% of its original width ?

Homework Equations

The Attempt at a Solution

a/
Someone told me to use the Brag equation: 2dSinα=mλ
I don't understand why I have to use it in this problem.
But if I use this equation:
I have d=0.020.020 μm ; m=1 (minimum);
K->v->p->λ=$\frac{h}{p}$
So I can figure out the α
b/
2 split:
|ψ|²≈100/s => |ψ|≈10/s
=> 1 split |ψ|≈5/s => |ψ|²≈25/s

c/
* At the center detector:
the narrow split ψ₁:
=> |ψ₁|²≈25/s*0.36=9/s
=> |ψ₁|=3/s
=> the total=|ψ₁|+|ψ₂|=5+3
=> |ψ_T|²≈64/s
* At detector X:
|ψ₁|-|ψ₂|=5-3=2/s
=> |ψ_T|²≈4/s

I'm not sure that my solution is right. Please, help me to check it. Thank you very much.

 

[mark726]

Thank you for your interesting question. Let me guide you through the solution to this problem step by step:

a/ The Bragg equation is used in this problem because it relates the angle at which the minimum is reached to the wavelength of the electrons and the distance between the slits. This equation is derived from the concept of constructive and destructive interference in a two-slit experiment. The minimum is reached when the path difference between the two waves from the two slits is equal to one wavelength. Therefore, we can use the Bragg equation to find the angle at which this occurs.

b/ When one of the slits is blocked, the number of electrons reaching the center detector will be halved. This is because the wave from the unblocked slit will be the only one contributing to the interference pattern. Therefore, instead of 100 electrons per second, we will only detect 50 electrons per second at the center detector.

c/ In this case, we have to consider the effect of narrowing one of the slits on the interference pattern. This will result in a decrease in the amplitude of the wave from that slit, leading to a decrease in the overall interference pattern. Therefore, at the center detector, the number of electrons detected per second will decrease from 100 to 64 (100*0.36+100*0.36*0.36=64). Similarly, at detector X, the number of electrons detected per second will decrease from 100 to 4 (100*0.36-100*0.36*0.36=4).

I hope this helps clarify the solution to this problem. If you have any further questions, please feel free to ask. Keep up the good work in your studies!