Wavelength - differences in equations

  • #1
Hi

If I wanted to find a wavelength of electrons having an average speed of 1.7e+8, would use lamdba = h/ mv or lamdba = hc/ E? I noticed that the 2nd eq is used mostly for transition states when an electron either falls or jumps from its n shell. What's the major difference?

Thanks in advance
 

Answers and Replies

  • #2
441
2
In the beginning of 20th century the interplay of special relativity of Einstein and the photon quantization hypothesis of Planck (that light is emitted or absorbed in parcels of energy E = h nu, called photons) lead to the formula

lambda = h / p

applicable to photons. Around 1922 de Broglie guessed that material particles like electrons may also have a wave associated with them so he postulated the same exact formula applies to electrons and other material particles.

The moment of the mass zero relativistic photons is p = E/c which leads to lamdba = hc/ E.
The moment of non-relativistic electrons is p = mv, leading to lamdba = h/mv.

Your two formulas are just two particular cases of the same master formula.
 
Last edited:
  • #3
jtbell
Mentor
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If I wanted to find a wavelength of electrons having an average speed of 1.7e+8,
I assume the speed is in m/sec.

would use lamdba = h/ mv or lamdba = hc/ E?
Neither one. The correct starting point for the wavelength is [itex]\lambda = h / p[/itex] where p is the momentum of the particle.

[itex]\lambda = h/mv[/itex] uses the non-relativistic momentum [itex]p=mv[/itex] instead of the relativistic momentum [itex]p = mv / \sqrt {1 - v^2/c^2}[/itex]. Your speed is more than half the speed of light, so it makes a significant difference.

[tex]\lambda = hc/E[/itex] applies only to massless particles like photons, for which [itex]E = pc[/itex], that is, [itex]p = E/c[/itex].
 
  • #4
Thanks guys

I'm also wondering about what electron transitions correspond to a UV light coming out from a hydrogen atom. How do I calculate that? And how do I differentiate that from the visible light?
 

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