Question

A photon of wavelength 2.0 x 10^–11 m strikes a free electron of mass m_e that is initially at rest, as shown direction, above left. After the collision, the photon is shifted in wavelength by an amount Δλ = 2h/m_ec, and reversed in direction, as shown above right.

(a) Determine the energy in joules of the incident photon.

(b) Determine the magnitude of the momentum of the incident photon.

(c) Indicate below whether the photon wavelength is increased or decreased by the interaction.

.................Increased ................ Decreased Explain your reasoning.

(d) Determine the magnitude of the momentum acquired by the electron.

Answer 1

(a) E = hc / λ … (1.99 x 10^-25 J-m) / 2x10^–11 … E = 9.9x10^–15 J

(b) p = h / λ … 6.63x10^–34 / 2x10^–11 … 3.3x10^–23 kg-m/s

(c) Due to energy conservation, the total energy before must equal the total energy after. Since some of the energy after is given to the electron, the new photon would have less energy than the original. From E = hc / λ, less energy would mean a larger λ.

(d) First determine the λ of the new photon. λ_new = λ_old + Δ λ … with Δ λ given in the problem as 2h/m_ec

λ_new = 2x10^–11 + (2*6.63x10^–34/(9.11x10^–31*3x10⁸)) … λ_new = 2.5x10^–11 m

Using momentum conservation

p_photon(i) = – p_photon(new) + p_electron (the new photon has – momentum since it moves in the opposite direction).

From this equation we get … p_electron = p_photon(i) + p_photon(new)  (with p = h/ λ for each photon)

P_electron = (h/ λ_i + h/ λ_new) = 6.63x10^–34 (1 / 2x10^–11 + 1 / 2.5x10^–11) … = 6x10^–23 kg-m/s