Question

Question #2e3c2

Answer 1

Here's what I got.

Explanation:

Actually, all you have to do here is convert the frequency of the photon to wavelength by using the equation

`color(blue)(ul(color(black)(lamda * nu = c)))`

Here

• `lamda` is the wavelength of the photon
• `nu` is its frequency
• `c` is the speed of light in a vacuum, usually given as `3.00 * 10^8"m s"^(-1)`

Keep in mind that you have

`"1 MHz" = 10^6"Hz" " "` and `" ""1 Hz" = "1 s"^(-1)`

which means that the frequency of the photon is equal to

`0.15 color(red)(cancel(color(black)("MHz"))) * (10^6color(red)(cancel(color(black)("Hz"))))/(1color(red)(cancel(color(black)("MHz")))) * "1 s"^(-1)/(1color(red)(cancel(color(black)("Hz")))) = 1.5 * 10^5"s"^(-1)`

Rearrange the equation to solve for `lamda`

`lamda * nu = c implies lamda = c/(nu)`

Plug in your value to get

`lamda = (3.00 * 10^8 "m" color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^6color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(2.0 * 10^2"m")))`

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

Now, the second equation you listed

`color(blue)(ul(color(black)(E = h * nu)))`

where

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

is called the Planck - Einstein relation and can be used to find the energy of a photon that has `nu` as its frequency.

You don't need to use this equation in order to find the wavelength of the photon, but you can rewrite it as a function of wavelength

`E = h * c/(lamda)`

This means that the wavelength of the photon can be written in terms of its wavelength as

`lamda = (h * c)/E`