# Wavelength - differences in equations

1. Feb 19, 2008

### 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?

2. Feb 19, 2008

### 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.

Last edited: Feb 19, 2008
3. Feb 19, 2008

### Staff: Mentor

I assume the speed is in m/sec.

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, $p = E/c$.

4. Feb 20, 2008

### 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?

5. Feb 20, 2008