Wavelength of Light: Frequency & Photons Explained

[Someone502]

is the wavelength of light the frequency that it sends out photons? i don't quite get it thanks

 

[James R]

For any wave, not just light:

$$v=f \lambda$$

where v is the speed of the wave (e.g. speed of light), f is the frequency and $\lambda$ is the wavelength.

The frequency is the number of wave "crests" which pass a particular point in space per second. The wavelength is the distance between two "crests" of the wave. It is probably easiest to think about these quantities with something familiar, like water waves.

The wavelength of light is the distance between two maximum values of the electric (or magnetic) field making up the light wave. The frequency is the number of these field maxima which pass a point in space in a particular time.

 

[The Bob]

is the wavelength of light the frequency that it sends out photons? i don't quite get it thanks

I think the question you are asking needs two equations.

$$c = \nu \lambda$$ and $$E = h \nu$$

Light and all other elctromagnetic radiation can be explained in two forms: as a wave and as a particle.

As a wave, electromagnetic radiation (which includes radio waves, micro waves, infrared rays, visible light, ultra-violet light, x-rays and gamma rays) has a wavelength (denoted by Lambda) and a frequency (denoted by Nu). This relationship is what is represented by the first equation, above, (when c is the speed of light).

As a particle, electromagnetic radiation produces quanta (in the case of visible light these are known as photons). Each quantum has so much energy , which is related by the frequency of the quantum. This relationship is the second equation, above, (when h is Planck Constant).

So we can find out what the wavelength of light with a frequency of 4.5 x 10¹⁴ Hz is:

$$c = \nu \lambda$$

$$\Rightarrow 3.0 \times10^8 \ ms^{-1} = 4.5 \times 10^{14} \ Hz \ \times \ \lambda$$

$$\Rightarrow \lambda = \frac{3.0 \times10^8}{4.5 \times 10^{14}} = 6.67 \times 10^{-7} \ m$$

Alternatively, we can find out the energy that must be absorbed by an atom to produce a quantum with a frequency of 4.5 x 10¹⁴ Hz:

$$E = h \nu$$

$$\Rightarrow E = 6.63 \times 10^{-34} \ J Hz^{-1} \ \times \ 4.5 \times 10^{14} = 2.98 \times 10^{-19} \ J$$

So once 2.98 x 10^-19 Joules of energy has been absorbed by an atom, the energy then released will produce a photon with a frequency of 4.5 x 10¹⁴ Hz.

What you must remember is that the two ideas of light are different: particles and waves. Each need a separate equation but they are both linked by the frequency of the wave/photon.

Your original question was (to me) saying that the wavelength of light is the same as the frequency. I hope you now see that it is related by not the same (value).

The Bob (2004 ©)