Question

A proton and alpha particle are acclerated through the same accelerating potential. Which one of the two has (a) greater value of de-broglie wavelength associated with it, and (b) less kinetic energy? justify your answer.

Answer 1

(i) When a charged particle of charge q mass m, is accelerated under a pot. Diff. V, let v be the velocity acquired by the particle, then
`qV=1/2 mv^2 or mv=[2mqV]^(1//2)`
`:. lambda=h/(mv) =h/(sqrt(2mqV)) or lambda prop 1/(sqrt(mq))`
Hence, `(lambda_p)/(lambda_(alpha))=sqrt((m_(alpha)q_(alpha))/(m_pq_p))=sqrt((4m)/mxx(2e)/e)=2sqrt2gt1`
So, `lambda_pgtlambda_(alpha)`, i.e., de-Broglie wavelength associated with proton is greater than that of alpha particle.
(ii) Kinetic energy of charged particle
`E_k=qV, i.e., E_kpropq`
`:. (E_(k_p))/(E_(k_h))=(q_p)/(q_(alpha))=e/(2e)=1/2 lt 1` or `E_(k_p) lt E_(k_(alpha))`
i.e., proton particle has less kinetic energy than `alpha`-particle.