Question

Question #15e99

Answer 1

The IR photon.

Explanation:

The energy of a photon is directly proportional to its frequency as described by the Planck - Einstein relation

`color(blue)(|bar(ul(color(white)(a/a)E = h * nu color(white)(a/a)|)))`

Here

`E` - the energy of the photon
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`
`nu` - the frequency of the photon

For two photons of frequency `nu_1` and `nu_2`, you can say that

`E_1 = h * nu_1" "` and `" "E_2 = h * nu_2`

If you were to divide these two equations, you would find

`E_1/E_2 = (color(red)(cancel(color(black)(h))) * nu_1)/(color(red)(cancel(color(black)(h))) * nu_2)`

Rearrange to get

`E_2 = nu_2/nu_1 * E_1`

If `nu_2>nu_1`, i.e. if the first photon has a higher frequency than the second photon, then `E_2 > E_1`, meaning that it's also more energetic.

If `nu_2 < nu_1`, i.e. if the first photon has a lower frequency than the second photon, then `E_2 < E_1`, meaning that it's also less energetic.

Now, take a a look at the EM Spectrum.

Notice that infrared frequency measures at about `10^12"Hz"` and microwave frequency measures at about `10^9"Hz"`.

Since a photon of IR radiation has a higher frequency than a photon of microwave radiation, it follows that it will also be more energetic.

In fact, you can say that you have

`E_"IR" = (10^12 color(red)(cancel(color(black)("Hz"))))/(10^9color(red)(cancel(color(black)("Hz")))) * E_"microwave"`

`E_ "IR"= 10^3 * E_"microwave"`

Therefore, a photon of frequency `10^12 "Hz"` is about `10^3` times more energetic than a photon of frequency `10^9"Hz"`.