Question

A Proton and − particle have the same De-Broglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds.

Answer 1

lambda=h/p=h/mv=h/√2mqV
here, v = accelerating potential & v = speed of particle

As it is given that A proton and a particle have same de-broglie wavelength

charge on proton=qp
mass on proton=mp
charge on alpha particle= 2qp
mass of alpha particle= 4mp

lambda2=lambdap

(next step is in attachment)

Answer 2

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