A photon of wavelength 2.0 x 10^-11 m strikes a free electron of mass me that is

Question

A photon of wavelength 2.0 x 10^-11 m strikes a free electron of mass me that is initially at rest, as shown above left. After the collision, the photon is shifted in wavelength by an amount ? = 2h/mec, and revered 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. Explain your reasoning.
(d) Determine the magnitude of the momentum acquired by the electron.

Explanation / Answer

(a) Energy of the photon is given by,

E = hc/

h = Planck constant = 6.626 x 10^-34 Js

c = speed of light = 3 x 10^8 m/s

= 2 x 10^-11 m

E = 6.626e-34*3e8/2e-11 = 9.94 x 10^-15 J

(b) Momentum of the photon is given as,

p = h/ = 6.626e-34/2e-11 = 3.31 x 10^-23 kg.m/s

(c) Photon will lose energy in the collision, so

E2 < E1

hc/2 < hc/1

2 > 1

Thus, the wavelength will increase.

(d) 2 = 1 + 2h/(me*c)

me = 9.1 x 10^-31 kg

2 = 2e-11 + 2*6.626e-34/(9.1e-31*3e8) = 2.49 x 10^-11 m

The magnitude of the momentum of the photon will be,

p2 = h/2 = 6.626e-34/2.49e-11 = 2.66 x 10^-23 kg.m/s

The magnitude of the momentum of the electron will be

pe = p1 - p2 = 3.31e-23 - 2.66e-23 = 6.5 x 10^-24 kg.m/s

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