What does omega mean in the photon energy derivation?

[Randomchemstudent]

Homework Statement

w/v=c,w=vc giving E=hcv

Relevant Equations

E=hw, lamda*w=c, wavenumber (v) = 1/lamda

So I'm a chemistry student here in the UK - and I'm feeling a bit like a fish out of water on a physics forum but... I'm having trouble with a derivation that seems simple but i can't for some reason seem to understand.

I know that the energy of a photon is given by E=hc and that c=lambda*v giving E=hc/lamda. But my lecturer has used w(omega) in his derivation and i don't understand what it means and why he's used it to derive the energy in this way.

Any help would be much appreciated!

 

[Orodruin]

$\omega$ is angular frequency. It differs from frequency by a factor of $2\pi$.

 

[mjc123]

Spectroscopists often use ω for frequency, or for wavenumber. They (we) often loosely talk about wavenumber as frequency, e.g. "a frequency of 1730 cm^-1". For example, on the NIST website, molecular vibrational frequencies in cm^-1 are given the symbol ω. So you have to be a bit careful.

 

[Herman Trivilino]

Problem Statement: w/v=c,w=vc giving E=hcv
Relevant Equations: E=hw, lamda*w=c, wavenumber (v) = 1/lamda

I know that the energy of a photon is given by E=hc

No. The energy of a photon is $hf$, which is often written as $h \nu$, where $f$ or $\nu$ is the frequency. You seem to have your notation mixed up.