a proton and an electron are accelerateed by same potential differnce  . Let l nd p denote the de Brogli wavelength of the electron and the proton respectivly

a) l = p

b)  l < p

c) l > p

 d) the relation depends on accelerating pot. deifference.

i think I=p

yes    w=qv is same for both .  so the net gain in K.Eof both is also same.

w=qv=1/2m₁v₁²                 1 for electron

w=qv=1/2m₂v₂²                   2 for proton

from above m₁ / m₂=(v₂ / v₂)^{2....................................}A

the de Brogli eqn for e  is I=h  /  m₁v_{1.....................}  1

the de Brogli eqn for proton    p=h/m₂v_{2 ...................} 2

 from 1 and 2

                I/p=m₂v₂ / m₁v₁................................B

from eqn A   v₂ / v₁=(m₁ / m₂ )^{1/2....................................}C

putting C in B

                           I/p=(m₂  / m₁)^1/2

                         as m₂ >m₁

                               I>p