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An electron and a proton are moving in the same direction and possess same kinetic energy. Find the ratio of de-Broglie wavelengths associated with these particles. |
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Answer» <P> Solution :We KNOW that de-Broglie wavelength `lamda=(h)/(MV)=(h)/(sqrt(2mK))`, where K is the kinetic energy. As `m_(p)gtm_(e)`, hence it is clear that for same kinetic energy.`(lamda_(e))/(lamda_(p))=sqrt((m_(p))/(m_(e)))`. `implies lamda_(e) lamda_(p)` i.e, de-Broglie wavelength of electron will be GREATER than that of PROTON. |
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