Uncertainty Principle and single-slit diffraction

In summary: You should get a value of θ, then you multiply by 2D to get the width of the screen pattern in meters.In summary, By using the Heisenberg uncertainty principle and assuming the vertical positions of the electrons are confined within the slit, we can estimate the minimum uncertainty in the y-component of their momentum after passing through the slit. This can then be used to estimate the width of the diffraction pattern produced on the screen. The formula for this is d sinθ = λ, where d is the width of the slit and θ is the angle between the center of the screen and either first minimum on that screen. By using the small angle approximation and ΔyΔpy = h, we can relate θ
  • #1
throneoo
126
2

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
 
Last edited:
Physics news on Phys.org
  • #2
throneoo said:

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
rude man said:
ΔyΔpy ≥ h
thanks for the guidance. but i still cannot convince myself about this. how do we find out what the constant is?
 
  • #4
throneoo said:
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).
 
Last edited:
  • #5
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
ok so apparently you want \hbar and not \bar{h} in latex
 
  • #7
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.
 
  • Like
Likes throneoo
  • #8
rude man said:
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
throneoo said:
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.
 

What is the Uncertainty Principle?

The Uncertainty Principle, also known as Heisenberg's Uncertainty Principle, states that it is impossible to simultaneously know the exact position and momentum of a subatomic particle. This is due to the fact that the act of measuring one of these properties will inevitably change the other, making it impossible to obtain precise measurements for both.

How does the Uncertainty Principle relate to single-slit diffraction?

The Uncertainty Principle plays a role in single-slit diffraction because it limits our ability to accurately measure both the position and momentum of particles as they pass through the slit. This leads to a spread in the particle's momentum, which manifests as a diffraction pattern on the other side of the slit.

What is single-slit diffraction?

Single-slit diffraction is a phenomenon that occurs when a beam of particles, such as light, passes through a narrow opening or slit. The particles are diffracted, or spread out, as they pass through the slit, resulting in a characteristic pattern on the other side.

How does the width of the slit affect single-slit diffraction?

The width of the slit has a direct impact on the diffraction pattern produced. A narrower slit will result in a wider diffraction pattern, while a wider slit will produce a narrower diffraction pattern. This is because a wider slit allows for more particles to pass through, resulting in a more concentrated diffraction pattern.

What are some real-world applications of the Uncertainty Principle and single-slit diffraction?

The Uncertainty Principle and single-slit diffraction have many practical applications, such as in the field of nanotechnology where precise measurements of subatomic particles are necessary. They also have applications in various imaging techniques, such as electron microscopy and medical imaging. Additionally, the principles are used in the development of new materials and technologies, such as diffraction gratings and optical filters.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
5K
Replies
2
Views
175
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
8K
  • Introductory Physics Homework Help
Replies
10
Views
8K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
57
Views
4K
Back
Top