Question

Question #9af62

Answer 1

Here's what I got.

Explanation:

The thing to remember about frequency and wavelength is that they have an inverse relationship described by 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 * 10^(8)` `"m s"^(-1)`

In your case, the wavelength of the photons is given in nanometers, so start by converting it to meters

`545 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 5.45 * 10^(-7)color(white)(.)"m"`

Rearrange the equation to find the frequency of the photons

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

Plug in your value to find

`nu = (3 * 10^8 color(red)(cancel(color(black)("m"))) "s"^(-1))/(5.45 * 10^(-7)color(red)(cancel(color(black)("m"))))`

`color(darkgreen)(ul(color(black)(nu = 5.50 * 10^(14)color(white)(.)"s"^(-1))))`

The answer is rounded to three sig figs, the number of sig figs you have for the wavelength of the photons.

Now, to find the energy of a photon of wavelength `'545 nm"`, you need to use the Planck - Einstein relation, which looks like this

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

Here

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

Plug in your values to find

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 5.50 * 10^(14) color(red)(cancel(color(black)("s"^(-1))))`

`color(darkgreen)(ul(color(black)(E = 3.64 * 10^(-19)color(white)(.)"J")))`

Once again, the answer is rounded to three sig figs.