A photon of 680-THz light strikes a stationary atomic particle of mass 8.9 × 10^

Question

A photon of 680-THz light strikes a stationary atomic particle of mass 8.9 × 10^-26 kg.

(a) Assuming the photon is absorbed, find the resulting speed of the particle, in meters per second.

(b) Now assume that, instead of being absorbed, the photon reflects off the particle at an angle of 5.3° away from the photon’s incident direction. Assume the photon’s frequency remains the same, although that would not be strictly correct. Find the resulting speed of the particle, in meters per second.

(c) An atomic system might absorb an incident photon and then emit a second photon. If the particle of part (a), after absorbing the photon as described there, emits a photon of 440-THz light in the opposite direction to the first photon, find the final speed of the particle, in meters per second.

Explanation / Answer

a)frequency of light (f)=680x10^12hz

mass of stationary particle(m)=8.9x10^-26kg

a)photon is absorbed by stationary particle then applying law of conservation of energy we get

hf=(1/2)mv^2

v=sqrt((2hf)/m)

=((2x6.63x10^-34x680x10^12)/(8.9x10^-26))^(1/2)

=3183m/s

b)photon reflects with 5.3degree with same frequency .so according to law of conservation of energy the velocity of the particle is zero.

c)Again applying the law of conservation of energy we get

hf=hf1+(1/2)mv^2

v=((2h(f-f1))/m)^(1/2)

here f=680THz

f1=440THz

substituting the values we get

v=((2x6.63x10^-34(680-440)10^12)/(8.9x10^-26))^0.5

v=1890.96m/s

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