Wavelength - differences in equations

[The_ArtofScience]

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

 

[smallphi]

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.

 

[jtbell]

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 $\lambda = h / p$ where p is the momentum of the particle.

$\lambda = h/mv$ uses the non-relativistic momentum $p=mv$ instead of the relativistic momentum $p = mv / \sqrt {1 - v^2/c^2}$. 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 $E = pc$, that is, $p = E/c$.

 

[The_ArtofScience]

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?

 

[smallphi]

http://en.wikipedia.org/wiki/Uv_light
http://en.wikipedia.org/wiki/Rydberg_formula