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A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?

Answer»

Solution :As `lamda=(h)/(sqrt((2mK)))` and kinetic energy K is CONSTANT, HENCE `lamdaprop(1)/(sqrtm)` i.E., electron particle, being LIGHTER, will have the larger de-Broglie wavelength. Thus `lamda_(e) GT lamda_(p)`.


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