Question

Question #e56e4

Answer 1

`6.6 * 10^(-21)` `"J"`

Explanation:

For starters, the frequency you have here is way too low to match that of a gamma ray. Gamma rays have frequencies that are `> 10^(20` `"Hz"`.

In fact, the frequency given to you matches that of a photon in the infrared portion of the EM spectrum.

I'll show you hot to find the energy of this photon, but keep in mind that you are not dealing with a gamma ray here!

The most important thing to remember about the energy of a photon is that it is directly proportional to its frequency as described by the Planck - Einstein relation.

`E =h * nu`

Here

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

This basically means that a photon that has a high frequency will also have a high energy. In other words, the higher the frequency of the photon, the higher its energy.

Now, notice that the frequency of the photon is given to you in hertz, `"Hz"`. In order to make the units consistent with the Planck - Einstein relation, you can express the frequency in `"s"^(-1)` by using the fact that

`"1 Hz" = "1 s"^(-1)`

Plug in the value you have for the frequency of the photon and solve for `E`.

`E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 1.0 * 10^(13)color(red)(cancel(color(black)("s"^(-1))))`

`E = color(darkgreen)(ul(color(black)(6.6 * 10^(-21)color(white)(.)"J")))`

The answer is rounded to two sig figs, the number of sig figs you have for the frequency of the photon.