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A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has (1) greater value of de-Broglie wavelength,

associated with it and (2) less kinetic energy? Explain.


(a) de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that λ∝1(mass)(charge) Mass and charge of a deuteron are 2mp and e respectively, and, mass and charge of an alpha particle are 4mp and 2e respectively. where, mp is the mass of a proton and e is the charge of an electron λDλα=mαqαmDqD=(4mp)(2e)(2mp)(e)=21 Thus, de-broglie wavelength associated with deutron is twice of the de-broglie wavelength of alpha particle.

(b) K.E ∝q (for same accelerating potential) Charge of a deuteron is less as compared to an alpha particle. So, deuteron will have less value of K.E.
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