Does the de Brogile wavelength of an electron (*lambda*=h/p) equal v/f as a normal wavelength does?

Question

.sam. · Answer

pretty sure its related, with v=c
c/f=h/p
cp=hf
E=hf

anonymous · Answer

i m not sure.the derivation is fine but can matter waves have such energy as hf...i think that part should be applied to only photons..matter waves should be having kinetic energy instead..

.sam. · Answer

i'm not sure too, but this chart seems to be right

anonymous · Answer

Okay, thanks!

.sam. · Answer

Some info http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html#c1

anonymous · Answer

hmmm..thanks for the link..the chart says nothing about the frequency though ,i observed..

anonymous · Answer

If it's any consolation I tried to use the relationship $$v \div f = h \div mv $$ to get the velocity  $$v ^{2}=hf \div m $$ and I got the wrong answer so I assume it doesn't work.

anonymous · Answer

i think somehow the energy expression E=hf is valid only for zero rest mass particles only..

.sam. · Answer

yes