The fomula h/lambda is this for the photon only?

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The formula h/λ applies specifically to massless particles, such as photons, and does not extend to relativistic electrons, which require different energy equations. For relativistic particles, the energy is expressed differently than the classical kinetic energy equation 1/2 mv², as relativistic effects must be considered. The discussion references de Broglie's work, clarifying that while his equation relates momentum and wavelength, it is applicable to massless particles. The equation E=fh, however, is valid for all particles, highlighting the quantization of energy. Overall, the distinction between massless and massive particles is crucial in understanding these relationships in physics.
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the fomula \frac{h}{\lambda}

is this for the photon only? or can it be applied to relativistic electrons too?
 
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it applies to particles with zero rest mass. Hence it won't apply to relativistic electrons.
 
so for relativistic electrons, if I wanted it's speed, i'd use .5mv^2?
 
1/2 m v^2 only works for non-relatavistic speeds, the energy for a relatavistic particle is different. See here for more details.
 
anjor said:
it applies to particles with zero rest mass. Hence it won't apply to relativistic electrons.

Same question was asked by de Brolie. And actually, it turned out that it will.
 
gulsen said:
Same question was asked by de Brolie. And actually, it turned out that it will.

Yes, but UrbanXrises' original formula was either a typo or assumed that c=1. With c=1 this formula is, in fact, only good for massless particles. DeBroglie's relationship involves the speed, which is less than c.

-Dan
 
No, it comes from:
E = pc = \frac{hc}{\lambda}
where c's cancel, and de Broglie's equation relates momentum and wavelength.
 
gulsen said:
No, it comes from:
E = pc = \frac{hc}{\lambda}
where c's cancel, and de Broglie's equation relates momentum and wavelength.

:redface: I was thinking of the energy equation. Sorry! (Ahem!)

Even though I got my c's wrong, the argument still holds...E=pc only hold for massless particles, which was what I was trying to say.

-Dan
 
Last edited:
so p=\frac{h}{\lambda} is for massless particles

but what about E=fh?

is this equation for massless particles too?
 
  • #10
UrbanXrisis said:
so p=\frac{h}{\lambda} is for massless particles

but what about E=fh?

is this equation for massless particles too?

No. This equation is good for anything. Basically this equation simply expresses the quantizability of energy.

-Dan
 
  • #11
UrbanXrisis said:
so p=\frac{h}{\lambda} is for massless particles

but what about E=fh?

is this equation for massless particles too?

No, it applies to all particles! That's the backbone for Schrödinger equation!
 

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