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Monochromatic light of wavelength 632.8nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam. (b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Asume 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 momentum a that of the photon ? |
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Answer» SOLUTION :Here `LAMDA=632.8nm=6.328xx10^(-7)m` and power emitted=`9.42mW=9.42xx10^(-3)W=9.42xx10^(-3)Js^(-1)` (a) Energy of each photon `E=(hc)/(lamda)=(6.63xx10^(-34)xx3xx10^(8))/(6.328xx10^(-7))=3.14xx10^(-19)J` and momentum of each photon `p=(E)/(c)=(3.14xx10^(-19))/(3xx10^(8))=1.05xx10^(-27)kg" "ms^(-1)` (B) Number of photons arriving per second at the target `n=(Power)/("Energy of one photon")=(9.42xx10^(-3))/(3.14xx10^(-19))=3xx10^(16)" photons "s^(-1)`. (c) We know that the mass of a hydrogen atom `m_(H)=1.67xx10^(-27)kg.` if velocity of hydrogen atom be V, then momentum of hydrogen atom `m_(H)v=`momentum of a photon=`1.05xx10^(-27)" ms"^(1)` `impliesv=(1.05xx10^(-27))/(m_(H))=(1.05xx10^(-27))/(1.67xx10^(-27))=0.63ms^(-1)`. |
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