Simple positron diffraction problem

In summary, the conversation discusses a problem where a beam of positrons travels at 0.001% the speed of light and impinges on a slit that is 1 μm wide. The question asks for the angle θ between the center line and 3rd minimum of the resulting diffraction pattern, as well as the distance to a detector for the 3rd minimum to be spaced 20 cm from the center line. The correct equations to use are nλ=asinθ and Y=Ltanθ, with a=1e-6 m and n=3. The wavelength is found to be 242.7 nm and the distance L is calculated to be 18.8339 cm.
  • #1
asdf12312
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1

Homework Statement



A beam of positrons (positron ≡ antielectron) travels at 0.001% the speed of light and impinges on a slit that is 1 μm wide. Use modern units to find the angle θ between the center line and 3rd minimum of the resulting diffraction pattern. How far away (in cm) would you locate a detector so that the 3rd minimum is spaced 20 cm from the center line?

Homework Equations


wavelength = hc/pc where hc=1240nm*ev
and i believe p=mv where mass is same as electron =9.1e-31 and so pc=mc2=0.511e6

there is example in my notes where pc=mv*c=mc2(v/c) was used to find wavelength if v was given in % of c. but this is where I get confused because in the notes 0.1%*c becomes 0.001 somehow (10-2?)

for minima: (n-0.5)*wavelength=d*sin(θ) where asked to find θ, d=1e-6m and n=3 for this problem.

The Attempt at a Solution


so.. after various incorrect attempts it give me what the answer should be but problem is I don't know how to get it :(

θ = 46.7200 Degrees
Tries 3/3
L= 18.8339 cm
Tries 3/3

I think i should do something like find wavelength using 1240/(0.511e6*0.001) and if I do I get 2.43 nm. but i think I am doing it wrong because I get wrong θ :(
 
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  • #2
asdf12312 said:
but this is where I get confused because in the notes 0.1%*c becomes 0.001 somehow (10-2?)
Sure.
100% = 1
10% = .1
1% = .01
and so on.
 
  • #3
You used the wrong equation to find the angle. This is a question about diffraction, not interference. Your equation for the minima should be nλ=asinθ, where a is the slit width (in this case a=1e-6 m) and n=3.
Your wavelength equation is correct. Wavelength should equal 242.7 nm.
Use Y=Ltanθ to find L.
20 cm is your vertical displacement from center line.
 
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1. What is a simple positron diffraction problem?

A simple positron diffraction problem involves the scattering of positrons, which are positively charged subatomic particles, off of a crystalline material. This process allows scientists to study the structure and properties of the material.

2. How is a simple positron diffraction problem solved?

To solve a simple positron diffraction problem, scientists use a technique called positron annihilation spectroscopy. This involves sending a beam of positrons through the material and measuring the energy of the emitted gamma rays to determine the structure of the material.

3. What are the applications of studying simple positron diffraction problems?

Studying simple positron diffraction problems can provide valuable information about the atomic and molecular structure of materials, which can be useful in fields such as materials science, chemistry, and physics. It can also help in the development of new materials for various applications.

4. What are the limitations of using simple positron diffraction to study materials?

One limitation of using simple positron diffraction is that it is only suitable for studying crystalline materials. It cannot provide information about amorphous or disordered materials. Additionally, the technique requires specialized equipment and expertise, making it less accessible for some researchers.

5. How does simple positron diffraction compare to other techniques for studying materials?

Simple positron diffraction is a complementary technique to other methods such as X-ray diffraction and neutron diffraction. It can provide unique information about the structure and properties of materials, making it a valuable tool in materials research. However, it also has its own limitations and may not be suitable for all types of materials.

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