What is the Energy of a Light Quantum in eV Given Wavelength in Nanometers?

[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

 

[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).

 

[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

 

[Andrew Mason]

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.

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

 

[ConceptuallyInept]

Is MKS just the same as SI units?

 

[DieCommie]

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

 

[Andrew Mason]

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

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