# How Do You Derive the Compton Equation?

• latitude
In summary, the conversation is about deriving the Compton equation from the equations for conservation of momentum and energy. The process involves manipulating equations and combining them to eliminate the variable \psi and solve for \lambda. There is a possible error in the equation Pe2, where a - should be a +.
latitude

## Homework Statement

Derive the compton equation.

## Homework Equations

$$\lambda$$ - $$\lambda$$ = h/ mc (1 - cos$$\theta$$)
E = hf = hc/$$\lambda$$

## The Attempt at a Solution

Okay, I'm sorry this is so long, I'll try and make it as concise as it is possible for a whole blather of random crap to be :]

Conservation of momentum components:
h/$$\lambda$$ = h/$$\lambda$$(cos$$\theta$$) + Pe(cos$$\psi$$)
0 = h/$$\lambda$$(sin$$\theta$$) - Pe(sin$$\psi$$)

After some combining, squaring, and the like (getting rid of $$\psi$$):
Pe2 = (h/$$\lambda$$)2 - (h/$$\lambda$$)2cos2$$\theta$$ + (h/$$\lambda$$)2sin2$$\theta$$ - (h/$$\lambda$$)(h/$$\lambda$$)cos$$\theta$$

E2 = p2c2 + ER2
So
P2 = (E2 - ER2)/c2

So I plug that into my momentum (I'm not going to write the righthand side of the equation while i show what I did w/ that)

(E2 - ER2)/c2 = ...
((hc/$$\lambda$$)2 - (mc2)2)/c2 = .
I tried to get rid of the denominator 'c'...
(h/$$\lambda$$)2 - m2c2 = ...

(m2c2$$\lambda$$)/h = $$\lambda$$/$$\lambda$$ - h/$$\lambda$$cos$$\theta$$

After some more fiddling I get to this:

$$\lambda$$ = h/m2c2 - (h/$$\lambda$$)(h/m2c2)cos$$\theta$$

It's kind of close but not really... I can write out all the steps I made if that is necessary, but I'm kind of hoping I made one nice, simple-to-fix error that is glaringly obvious to the more experienced :)

Thank you :)

Hi latitude!

(have a lambda: λ and a theta: θ and a psi: ψ )
latitude said:
Pe2 = (h/$$\lambda$$)2 - (h/$$\lambda$$)2cos2$$\theta$$ + (h/$$\lambda$$)2sin2$$\theta$$ - (h/$$\lambda$$)(h/$$\lambda$$)cos$$\theta$$

eugh

the first - should be a +

Hi there,

Don't worry, deriving equations can be tricky and it's common to get stuck. Let's take a look at the steps you've already done:

1. You started with conservation of momentum components, which is a good starting point.
2. You squared the equations and added them together, which is also a valid step.
3. You used the equation E = pc to substitute for p.
4. You tried to simplify the equation by getting rid of the denominator c, which is a good idea.

However, from there, I can see where you went off track. Let's take a closer look at your final equation:

\lambda = h/m2c2 - (h/\lambda)(h/m2c2)cos\theta

The first term on the right-hand side is correct, but the second term is not. Let's break it down:

(h/\lambda)(h/m2c2)cos\theta = (hc/\lambda)(h/m2c2)cos\theta

Notice that you have c and \lambda in the numerator, but in your previous steps, you were trying to get rid of them. Also, the second term should have a minus sign, not a plus sign.

Here's a hint to help you move forward: try substituting E = hc/\lambda into your equation and see what you get. Remember, you want to end up with an expression for \lambda` in terms of \lambda and \theta.

I hope this helps and good luck with your derivation!

## What is the Compton Effect?

The Compton Effect, also known as Compton Scattering, is a phenomenon in which an incoming photon of high energy collides with a stationary electron, resulting in the photon losing some of its energy and changing its wavelength. This effect was discovered by Arthur Compton in 1923 and is an important concept in the field of quantum mechanics.

## Can you explain the mathematical derivation of the Compton Effect?

Yes, the mathematical derivation of the Compton Effect involves the conservation of both energy and momentum. By considering the interaction between the photon and electron as an elastic collision, we can use these conservation laws to derive the final energy and wavelength of the scattered photon.

## What is the significance of the Compton Effect in modern science?

The Compton Effect has several important applications in modern science. It helps us understand the behavior of photons and electrons, and it also plays a crucial role in imaging techniques such as X-rays and gamma-ray spectroscopy. Additionally, the Compton Effect is used in particle accelerators to study subatomic particles and their properties.

## How does the Compton Effect relate to the wave-particle duality of light?

The Compton Effect is a clear demonstration of the wave-particle duality of light. It shows that light can behave both as a wave and as a particle, depending on the situation. In the Compton Effect, the photon behaves as a particle by colliding with the electron, while its change in wavelength indicates its wave-like nature.

## Are there any practical applications of the Compton Effect?

Yes, the Compton Effect has several practical applications in various fields. One of the most significant applications is in medical imaging, where X-rays are used to create images of the body's internal structures. It is also used in materials science to study the atomic and molecular structure of materials, and in astrophysics to understand the behavior of high-energy photons in space.

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