Uncertainty Principle and single-slit diffraction

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Homework Statement


A beam of 50eV electrons travel in the x direction towards a slit of width 6 micro metres which is parallel to the y direction. The diffraction pattern is observed on a screen 2 metres away.
Use the Heisenberg uncertainty principle to estimate the minimum uncertainty in the y-component of the electrons' momentum after passing through the slit . Use it to estimate the width of the pattern produced on the screen

Homework Equations


dsinθ=nλ

ΔyΔP≥h/4π

λ=h/sqrt(2mE)=h/p

Width=2Dtanθ

The Attempt at a Solution



The first thing I did was to find out the wavelength of the electrons .

Then , I assumed the electron's vertical positions are confined within the slit , so Δy=d

This is where I got stuck .

First of all , I'm not sure what's meant by the width of the pattern . The distance between the 1st order fringes / max. order fringes ? if it's the latter , since d>>λ , the max order n I obtained was 34570 , and the width would be 3106 m , which is ridiculously large . and the HUP would be of no use at all in both cases,since all i need is the angle.

Second of all , I'm not sure what ΔyΔP is limited to . I'm still new to this and I've seen variants of the formula with different limits :h/4π, h/2π , h , h/2 .....
which makes me wonder if the limits are derived and vary from situations to situations , with h/4π being the lowest limit in theory .

Any help would be appreciated
 
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Answers and Replies

  • #2
rude man
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Homework Statement


A beam of 50eV electrons travel in the x direction towards a slit of width 6 micro metres which is parallel to the y direction. The diffraction pattern is observed on a screen 2 metres away.
Use the Heisenberg uncertainty principle to estimate the minimum uncertainty in the y-component of the electrons' momentum after passing through the slit . Use it to estimate the width of the pattern produced on the screen

Homework Equations


dsinθ=nλ
Let n = 1 and approximate sinθ = θ
ΔyΔP≥h/4π
There is no hard & fast rule. For this problem, go with ΔyΔpy ≥ h


The Attempt at a Solution



The first thing I did was to find out the wavelength of the electrons .

Then , I assumed the electron's vertical positions are confined within the slit , so Δy=d
[/quote]
Correct. This is a vital assumption.
This is where I got stuck .

First of all , I'm not sure what's meant by the width of the pattern . The distance between the 1st order fringes / max. order fringes ? if it's the latter , since d>>λ , the max order n I obtained was 34570 , and the width would be 3106 m , which is ridiculously large . and the HUP would be of no use at all in both cases,since all i need is the angle.
The width of the pattern is the distance between the two first minima.
Hint: assume the uncertainty corresponds to the distance from the center to a first minimum.
Start with ΔyΔpy = h
Relate Δpy to m and vy. Pretty obvious ...
Relate wavelength to the x axis speed (I would leave out numbers. Call it vx). I believe you already did this.
You can now relate Δy, Δvy, vx and λ.
By geometry you can also relate vy and vx to your angle θ.
Finally you can relate θ to λ and Δy which is the formula derived by wave theory (except for the sinθ = θ approximation.
 
  • #3
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ΔyΔpy ≥ h
thanks for the guidance. but i still cannot convince myself about this. how do we find out what the constant is?
 
  • #4
rude man
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thanks for the guidance. but i still cannot convince myself about this. how do we find out what the constant is?
What constant are you talking about? h? You're supposed to START with the Heisenberg uncertainty principle which is defined in terms of the known constant h (as Max Planck formulated it well before Heisenberg's 1927 relation).
 
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  • #5
BiGyElLoWhAt
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The original equation was ##\Delta x \Delta p \geq h##
But fermilab (?) reported back uncertainties ##\Delta x \Delta p \geq \hbar/2 = h/4pi## I'm pretty sure it was fermilab, and I'm pretty sure it was hbar and not h, but my uncertainty must be greater than... cutting the circle here.
 
  • #6
BiGyElLoWhAt
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ok so apparently you want \hbar and not \bar{h} in latex
 
  • #7
rude man
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Heisenberg posited the uncertainty products to be "of the order of h". So whether yiou go with h or h-bar or h/2 makes no difference.
 
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  • #8
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Let n = 1 and approximate sinθ = θ
does θ here refer to the angle subtended by the two minima ?

because I normally use θ as the angle between the horizontal axis and the fringes

then in that case , dsinθ=λ/2 , which is the path difference between the destructively interfering waves .

anyway , using the small angle approximation and ΔyΔpy = h,

px=h/λ

2θ=Φ=λ/Δy = Δpy / px , where θ is the angle between the horizontal axis and one of the first minima and Φ is that between the two first minima

the 'width' = 2Dθ = DΦ , where D is the distance between the screen and the slit
 
  • #9
rude man
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does θ here refer to the angle subtended by the two minima ?

because I normally use θ as the angle between the horizontal axis and the fringes

then in that case , dsinθ=λ/2 , which is the path difference between the destructively interfering waves .
θ is the angle between the center of the screen and either firtst minimum on that screen. The formula for that is d sinθ = λ, not what you wrote,
where d is the width of the slit; in your case d = Δy.
Below you state " ... where θ is the angle between the horizontal axis and one of the first minima" which is the same angle θ as mine.
anyway , using the small angle approximation and ΔyΔpy = h,

px=h/λ

2θ=Φ=λ/Δy = Δpy / px , where θ is the angle between the horizontal axis and one of the first minima and Φ is that between the two first minima
Make that θ = λ/Δy = Δpy / px and you've got a deal.
Everything else looks fine.
 

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