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

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Homework Help Overview

The discussion revolves around the application of the formula \( \frac{h}{\lambda} \) and its relevance to photons and relativistic electrons. Participants explore the conditions under which this formula applies, particularly focusing on the characteristics of massless particles versus those with mass.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question whether the formula \( \frac{h}{\lambda} \) is exclusive to photons or if it can also apply to relativistic electrons. There are discussions about the energy equations relevant to different types of particles, including the implications of mass and speed on these formulas.

Discussion Status

Multiple interpretations of the formulas are being explored, with some participants asserting that \( \frac{h}{\lambda} \) applies only to massless particles, while others reference de Broglie's work to suggest broader applicability. There is an ongoing examination of the conditions under which various energy equations hold true.

Contextual Notes

Participants are navigating the nuances of relativistic versus non-relativistic physics, particularly in relation to mass and the implications for energy equations. The discussion reflects a mix of established physics principles and ongoing inquiry into their applications.

UrbanXrisis
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the fomula [tex]\frac{h}{\lambda}[/tex]

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:
[tex]E = pc = \frac{hc}{\lambda}[/tex]
where c's cancel, and de Broglie's equation relates momentum and wavelength.
 
gulsen said:
No, it comes from:
[tex]E = pc = \frac{hc}{\lambda}[/tex]
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 [tex]p=\frac{h}{\lambda}[/tex] is for massless particles

but what about [tex]E=fh[/tex]?

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

but what about [tex]E=fh[/tex]?

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 [tex]p=\frac{h}{\lambda}[/tex] is for massless particles

but what about [tex]E=fh[/tex]?

is this equation for massless particles too?

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

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