Sunday, February 13, 2011

Mastering Physics: Compton Scattering

Learning Goal: To understand the derivation of the Compton scattering formula.
In 1923, Arthur H. Compton conducted a series of experiments that demonstrated the particle-like nature of electromagnetic radiation. Compton studied the collision of photons from a uniform wavelength beam with a stationary target. He observed that the radiation that exited from the target showed two characteristic wavelengths, one at the original wavelength lambda and another at a new wavelength lambda' shifted by an amount Delta lambda from the original. In this problem we will use the principles of conservation of energy and conservation of momentum, coupled with some basic ideas in quantum physics and relativity, to derive a statement for the shifted wavelength lambda'.

Part A
Consider a photon with initial momentum p_i on a collision course with a stationary electron of mass m. What is the total energy E_i of these two particles before the collision? Don't forget to include the rest mass energy of the electron.
Express your answer in terms of the p_i, the speed of light c, and m.
ANSWER:

  E_i  = p_{i}c+mc^{2}

Part B
After the collision, the photon exits with a momentum p_f at an angle phi from its initial momentum vector. The electron scatters off with a momentum p_e. What is the total energy after the collision? In this case, do not forget to include the relativistic energy of a particle.
Express your answer in terms of p_f, p_e, m, and c.
ANSWER:

  E_f  = p_{f}c+\sqrt{\left(mc^{2}\right)^{2}+\left(p_{e}c\right)^{2}}
Part C
The conservation of energy requires that the energy of the system before the collision E_i and the energy after the collision E_f be equal to one another. We want to manipulate this conservation equation to arrive at the result for Compton scattering. The first step is to take E_i as one side of the equation and E_f as the other side of the equation. Divide both sides by c and isolate the square root on one side by subtracting the term p_f from both sides of the equation. You should now be left with a trinomial on one side. Complete the right-hand side of the equation with these three terms.
Express your answer in terms of p_i, p_f, m, and c.
ANSWER:

  \sqrt{(mc)^2 + p_e^2}  = p_{i}-p_{f}+mc
Part D
Square both sides of the equation obtained in Part C and choose the result of squaring the right-hand side (the one you entered in the last part) as your answer. Your unreduced answer should have nine terms in it. If you further reduce your answer you should have at least six terms.
ANSWER:
(mc)^2 + p_e^2 =
Part E
You can use conservation of momentum to eliminate the term p_e from the equation. If you recall that p_e_vec, p_i_vec, and p_f_vec are all vectors you can use vector addition to state one of the vectors in terms of the other two. Give the value for p_e_vec in terms of p_i_vec and p_f_vec.
Express your answer in terms of p_i_vec and p_f_vec.
ANSWER:

  p_e_vec  = \vec{p_{i}}-\vec{p_{f}}
 
Part F
You now wish to get a value for p_e^2 in terms of the scalar quantities p_i, p_f, and the angle phi between the two vectors. Recall that when squaring a vector it is necessary to use the dot product: p_e^2 = (\vec{p}_e)^2= \vec{p}_e \cdot \vec{p}_e.
Express your answer in terms of p_i, p_f, and phi.
ANSWER:

  p_e^2  = p_{i}{^{2}}+p_{f}{^{2}}-2p_{i}p_{f}{\cos}{\phi}


Part G
Substitute the value you obtained for p_e^2 in Part F into the equation obtained in Part D. Eliminate terms, and divide by 2 so that one side of the equation reads p_ip_f - p_ip_f \cos(\phi). Supply the other side of the equation in the answer box.
Express your answer in terms of p_i, p_f, m, and c.
ANSWER:

  p_ip_f - p_ip_f \cos(\phi)  = p_{i}mc-p_{f}mc


Part H
In this final step, multiply both parts of the equation in Part E by h/(mp_ip_fc). By recalling the relationship between photon wavelength and momentum, p = h/\lambda, you can now arrive at a statement for the shifted wavelength \lambda^{\prime} -
\lambda. Give the remaining part of the equation that summarizes the Compton scattering relationship.
Express your answer in terms of Planck's constant h, the speed of light c, the rest mass of the electron m, and the angle phi.
ANSWER:

  \lambda^{\prime} - \lambda  = \frac{h}{mc}\left(1-{\cos}{\phi}\right)

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