A Quick Question on Compton Scattering

[j-e_c]

1. A photon of wavelength 1.000nm is scattered by a muon initially at rest. What is the deflection angle of the photon if its wavelength after scattering is 1.001nm?

Homework Equations

$$\lambda$$$$^{'}$$ - $$\lambda$$ = $$\frac{h}{mc}$$ (1 - cos($$\vartheta$$)

Assume that the mass of the muon is 200 times that of an electron.

The Attempt at a Solution

I have got down to cos($$\vartheta$$) = -81.4. I'm not sure what to do now.

 

[Doc Al]

I have got down to cos($$\vartheta$$) = -81.4.

That's not good. Getting |cosθ| > 1 is a sure sign that you've made an error somewhere.

Redo your calculation. (Show what you plugged in where.)

 

[j-e_c]

Thanks for the quick reply.

Note - I'm going to change theta to x so I can avoid LaTeX complications. Similarly, I'm going to change lambda to a and lambda-prime to b.

First, I rearranged the equation to make cos(x) the subject:

(b - a) = h/mc (1 - cos(x))

(b - a)mc/h = 1 - cos(x)

1 - [(b - a)mc/h] = cos(x)

Now, I'll show you how the value came about:

b - a = (1.001 - 1.000)x10^-9
= 1.000x10^-12 (1)

mc/h = (200 x 9.11x10^-31 x 3x10^8) / (6.63x10^-34) = 8.244x10^13 (2)

Multiplying (1) and (2) I get 82.44

Then 1 - 82.44 = -81.44

Therefore cos(x) = -81.44

 

[Doc Al]

Your calculation looks OK to me. That tells us that the assumptions of the problem are not consistent with Compton scattering. You can't have the given change in wavelength when scattering off of a muon! (Either that's a trick question or I'm missing something basic. :uhh: I'll think about it.)

 

[paranormal]

Firstly, I would like to clarify the question. Is the muon initially at rest or is it moving? If it is moving, what is its initial velocity? This information is necessary to accurately calculate the deflection angle.

Assuming the muon is initially at rest, we can use the equation you have provided to solve for the deflection angle. Plugging in the values, we get:

1.001nm - 1.000nm = \frac{h}{(200m_e)c} (1 - cos(\vartheta))

0.001nm = \frac{h}{(200m_e)c} (1 - cos(\vartheta))

Solving for cos(\vartheta), we get:

cos(\vartheta) = 1 - \frac{0.001nm(200m_e)c}{h}

Using the value of Planck's constant (h = 6.626 x 10^-34 J.s) and the mass of an electron (m_e = 9.109 x 10^-31 kg), we get:

cos(\vartheta) = 1 - \frac{0.001nm(200)(9.109 x 10^-31 kg)(3 x 10^8 m/s)}{6.626 x 10^-34 J.s}

cos(\vartheta) = 1 - 0.003 x 10^-3 = 0.9997

Taking the inverse cosine, we get:

\vartheta = cos^{-1}(0.9997) = 0.025 degrees

Therefore, the deflection angle of the photon is approximately 0.025 degrees.