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Monochromatic light of wavelength 3000 Å is incident normally on a surface of area 4cm^(2)If the intensity of light is 150(m W)/(m^(2)), the number of photon being incident on this surface in one second is …… |
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Answer» Solution :`LAMBDA=3000Å =3xx10^(-7)m` `A=4cm^(2)=4xx10^(-4)m^(2)` `I=150(mW)/(m^(2))=150xx10^(-3)(W)/(m^(2))` `h=6.62xx10^(-34) Js` `"intensity" I=(E )/(Axxt)=(P)/(A)` `therefore P=IA=150xx10^(-3)xx4xx10^(-4)` If number of photon colliding on the surface are n,its energy =power `P=(NHC)/(lambda)` `therefore n=(plambda)/((hc))=(6xx10^(-5)xx3xx10^(-7))/(6.62xx10^(-34)xx3xx10^(8))` `therefore n=9xx10^(13)s^(-1)` |
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