Question

An electron and a proton are accelerated through the same potential. Which one of the two has (i)b greater value of de-Broglie wavelength associated with it and (ii) less momentum?justify your answer.

Answer 1

When a charged particle of charge q, mass m is accelerated under a potential difference V, Let v be the velocity acquired by particle. Then
`qV=1/2mv^2 or mv=sqrt(2qVm)`
(i) `lambda=h/(mv)=h/(sqrt(2qVm)) or lambda prop 1/(sqrt(qm))`
`:. (lambda_e)/(lambda_p)=sqrt((q_pm_p)/(q_em_e))=sqrt((exx1837m_e)/(exxm_e))gt1`
So, `lambda_egtlambda_p`, i.e., greater value of de-broglie wavelength is associated with electron as compared to proton.
(ii) Momentum of particle, `p=mv=sqrt(2qVm)`
`:. p prop sqrt(qm)`,
Hence `(p_e)/(p_p)=sqrt((q_em_e)/(q_pm_p))=sqrt(e/exx(m_e)/(1837m_e)) lt 1`
So, `p_e lt p_p`, i.e., lesser momentum is associated with electron as compared to proton.