A proton and deutron 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 momentum? Give reasons to justify your answer.

Explanation

(a) De-Broglei wavelength is given by
`lambda=h/(sqrt(2qVm)) or lambda prop 1/(sqrt(qm))`
`:. (lambda_p)/(lambda_d)=sqrt((q_dm_d)/(q_pm_p))=sqrt((m_d)/(m_p)) ( :' q_d=q_p)`
As `m_dgtm_p, so lambda_pgtlambda_d`
(b) Momentum of a particle, `p=h/lambda`
i.e., `p prop1/lambda :. (p_p)/(p_d) =(lambda_d)/(lambda_p) lt 1 or p_p lt p_d`