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There are two sources of light,each emitting with a power of 100W.One emits X-rays of wavelength 1 nm and the other visible light at 500 nm.Find the ratio of number of photons of X-rays to the photon of visible light of the given wavelength? |
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Answer» Solution :Power`(E_(N))/(t)=(nhf)/(t)=(nhc)/(lambda)` `therefore P=n.(hc)/(lambda)` (where n.=no.of photons EMITTED PER unit time) `therefore n.=((P)/(hc))lambda` `therefore n.prop lambda` (`because` Here P,h,c are CONSTANT) `therefore (n.1)/(n._(2))=(lambda_(1))/(lambda_(2))` `=(1nm)/(500 nm)` `(n.1)/(n.2)=(1)/(500)` |
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