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A proton and an alpha-particle are accelerated through the same potential, which one of the two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy. Give reasons to support your answer. |
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Answer» SOLUTION :We know de-Broglie wavelength associated with a particle of CHARGE q, mass m, accelerated through a potential V is given by `lamda=(h)/(sqrt(2qmV))`. (i) Here accelerating potential V is CONSTANT for proton as well as `alpha`-particle. Hence `lamdaprop(1)/(sqrt(QM))`. As value of .qm. is less for proton, hence de-broglie wavelength for proton is more. (ii) The kinetic energy of a charged particle K=qV. as charged of proton is less than that of `alpha`-particle, the kinetic energy of proton is less. |
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