# Energy of a light quantum

1. Jun 2, 2005

### Jchem

I'm looking for an expression for the "energy of a light-quantum in eV when the wavelength is in nanometers"

and I'm kind of stumped

anyone know this formula?

thanks

2. Jun 2, 2005

### Nylex

You can use $E = \frac{hc}{\lambda}$, where h is Planck's constant, c is the speed of light and $\lambda$ is the wavelength (in metres). This will give you an answer in Joules, all you need to do then is convert to eV (1 eV = 1.6 x 10^-19 J).

3. Jun 2, 2005

### Jchem

ok so I can use $E = \frac{hc}{\lambda}$ X 1eV/1.6 x 10^-19 J

What about the "in nanometers" part?

The formula will work with a wavelength in nanometers, it will also work with any other size wavelengths..

not sure what they are asking here.

thanks

4. Jun 2, 2005

### Andrew Mason

The formula applies regardless of the units. You just have to use consistent units. If you use MKS, the energy is measured in Joules (m^2kg/sec^2), c is in m/sec, distance in m and h in Jsec. A nanometer is $10^{-9} metres$.

AM

5. Jun 5, 2005

### ConceptuallyInept

Is MKS just the same as SI units?

6. Jun 5, 2005

### DieCommie

I think MKS is meter-kelvin-seconds which is basically SI units.

7. Jun 5, 2005

### Andrew Mason

MKS is metre-kilogram-seconds or SI units (Système International d'Unités) as opposed to CGS (=centimetre-gram-seconds) or FPS (=foot pound seconds).

AM