We use the de Broglie expression:
#sf(lambda=h/(mv))#
So we can write:
#sf(lambda_(e)=h/(m_(e)v_(e))" "color(red)((1)))#
#sf(lambda_(p)=h/(m_(p)v_(p))" "color(red)((2)))#
#sf(v_(p)=3v_(e)" "color(red)((3))#
Substituting #sf(color(red)((3)))# into #sf(color(red)((2))rArr)#
#sf(lambda_(p)=h/(m_(p)3v_(e))" "color(red)((4)))#
Dividing #sf(color(red)((4)))# by #sf(color(red)((3))rArr)#
#sf(lambda_(p)/lambda_(e)=cancel(h)/(m_(p)3cancel(v_(e))).(m_(e)cancel(v_(e)))/(cancel(h))=1.8xx10^(-4))#
#:.##sf(m_(e)/(3m_(p))=1.8xx10^(-4))#
#sf(m_(e)/m_(p)=5.4xx10^(-4))#
This ratio is #sf(5.39522xx10^(-4))# for a neutron and #sf(5.36099xx10^(4))# for a proton, so a neutron seems to be the most likely candidate for the particle in question, though I guess there could be others.
