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Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon). |
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Answer» Solution :For PHOTON de-Broglie wavelength `lambda=(h)/(p)` , Momentum of electromagnetic RADIATION having v frequency and `lambda` wavelength, `p=(e )/(C )=(hv)/(c )` `THEREFORE p=(h)/(c ).(c )/(lambda.)[because v=(c )/(lambda.)]` `therefore p=(h)/(lambda.)` `therefore lambda.=(h)/(p)` but `(h)/(p)=lambda` de-Broglie wavelength of photon `therefore lambda.=lambda` Thus `lambda.` is wavelength of electromagnetic radiation which is equl to de-Broglie wavelength of phton with wavelength `lambda`. |
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