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Monichromatic light of wavelength 632.8 nm Is produced by a helium-neon laser.The power emitted is 9.42 mW. (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),and (c )How fast does a hydrogen atom have to travel in order to have the same moementum as that of the photon? |
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Answer» SOLUTION :Here `lambda=632.8 nm=632.8xx10^(-9)m` `P=9.42 mW=9.42xx10^(-3)W` `h=6.63xx10^(-34)Js,c=3xx10^(8) ms^(-1)` (a)energy of each photon, `E=(hc)/(lambda)` `therefore E=(6.63xx10^(-34)xx3xx10^(8))/(632.8xx10^(-9))` `=0.031431xx10^(-17)J` `therefore E~~3.14xx10^(-19)J` `therefore` Momentum of each photon, `impliesp=(h)/(lambda)` `therefore p=(6.63xx10^(-34))/(632.8xx10^(-9))` `therefore p=0.010477xx10^(-25)` `therefore p~~1.05xx10^(-27) kg ms^(-1)` (b)Let no. of photon reaching to target envery SECOND =N P=(no. of photon N) `P=NExx("energy of each photon E")` `therefore N=(P)/(E)=(9.42xx10^(-3))/(3.14xx10^(-19))` `therefore N=3x10^(16)"photon//second"` (c )Momentum of hydrogen atom=momentum of photon, `therefore` mv=p `therefore v=(p)/(m)=(1.05xx10^(-27))/(1.67xx10^(-27))` `therefore v=0.62874 therefore v~~0.63 m^(-1)` |
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