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Why are de-Broglie waves associated with a moving football not visible ? The wavelength lamda of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of the photon is (2lamdamc)/(h) times the kinetic energy of the electron, where m, c and h have their usual meanings. |
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Answer» Solution :de-Broglie waves associated with a MOVING football are not visible because in accordance with formula `lamda=(h)/(MV)`, the value of wavelength is extremely small (of the order of `10^(-34)`m). Energy of a photon of wavelength `lamda, E_(p)=hv=(hc)/(lamda)` if de-Broglie wavelength of electron be `lamda,` then `lamda=(h)/(mv) or V=(h)/(mlamda)` `THEREFORE` K.E. of an electron is `K_(e)=(1)/(2)mv^(2)=(1)/(2)m*((h)/(mlamda))^(2)=(h^(2))/(2mlamda^(2))`. . (ii) Dividing (i) by (ii), we have `(E_(p))/(K_(e))=(hc//lamda)/(h^(2)//2mlamda^(2))=(hc)/(lamda)xx(2mlamda^(2))/(h^(2))=(2mlamdac)/(h) or E_(p)=(2mlamdac)/(h)*K_(e)`. |
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