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6.4: The Compton Effect - Physics LibreTexts

Nov 5, 2020 ... The Compton effect is the term used for an unusual result observed when ... Describe how experiments with X-rays confirm the particle nature of radiation ... Here the photon's energy Ef is the same as that of a light quantum of ... The wave relation that connects frequency f with wavelength λ and speed c ... Teja
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Wavelength of monochromatic light \lambda = 632.8nm = 632.8 * 10^{-9} m

Power emitted P = 9.42 mW = 9.42 * 10^{-3} W

a) Find the energy and momentum of each photon in the light beam.

Energy of proton E=h\upsilon\ =\ \frac{hc}{\lambda}

Where, h - Planck's constant (6.626*10^{-34}), speed of light c = 3*10^{8} m/s   and \upsilon - frequency

E = \frac{6.626 *10^{-34} * 3*10^8}{632.8 * 10^{-9}}

E = 3.14*10^{-19} J

Momentum of each photon p = \frac{h}{\lambda}

p = \frac{6.626 *10^{-34}}{632.8 * 10^{-9}}

p = 1.047*10^{-27} kg m/s

(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area).

Relation between power and energy of photon

P = nE

n = \frac{P}{E}

n = \frac{9.42 * 10^{-3}}{3.14*10^{-19}}

n = 3.42 * 10^{16} potons/sec

c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Momentum  p = 1.047*10^{-27} kg m/s

Mass of hydrogen atom m = 1.66 * 10^{-27}

Momentum formula p = mv

v = \frac{p}{m}

Velocity   v = \frac{1.047*10^{-27}}{1.66 * 10^{-27}}

v = 0.621 m/s

Hence, Speed of hydrogen atom v = 0.621 m/s