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A photon of frequency v_i collides head-on with an electron of mass m moving with speed u_i. The photon is scattered in a direction exactly opposite to its initial direction of travel. It is observed that the frequency of the scattered photon equals the frequency of the incident photon. Which condition(s) must hold for this to occur?

1. The magnitude of the initial momentum of the electron is p_i = h*v_i / c.
2. The initial energy of the electron is E_i = h*v_i.
3. The magnitude of the initial momentum of the electron is p_i = 2*h*v_i / c.
4. The initial energy of the electron is E_i = (h*v_i)² / (2*m*c²).

Correct answer: The magnitude of the initial momentum of the electron is p_i = h*v_i / c.

Solution

Let the photon travel in the +x direction. Initial photon momentum = +h*v_i/c. Final photon momentum = -h*v_i/c (reversed). Momentum conservation: p_i^e + h*v_i/c = p_f^e - h*v_i/c => p_f^e = p_i^e + 2h*v_i/c. Energy conservation (same photon frequency): E_i^e = E_f^e. Since the electron energy is unchanged but its momentum changed, we need (p_f^e)² = (p_i^e)² (from E²=(pc)²+(mc²)² with same E). So p_f^e = -p_i^e. Then: -p_i^e = p_i^e + 2h*v_i/c => p_i^e = -h*v_i/c. Magnitude: |p_i^e| = h*v_i/c. (A) correct. KE = p²/(2m) = (h*v_i/c)²/(2m) = (h*v_i)²/(2mc²). (D) correct.

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