Question

Question #41bbe

Answer 1

Here's what I got.

Explanation:

The thing to remember about photons is that their energy is directly proportional to their frequency.

Simply put, the higher the frequency of a photon, the more energetic it will be. Likewise, the lower the frequency, the less energetic the photon.

This relationship between energy and frequency is described by the Planck - Einstein equation

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

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

Now, frequency and wavelength have an inverse relationship as described by the equation

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

Here

• `lamda` - the wavelength of the photon
• `c` - the speed of light in a vacuum, usually given as `3 * 10^8"m s"^(-1)`

So, you can now say that the photon with the shortest wavelength will have the highest frequency, which in turn will be equivalent to the highest energy.

Since

`lamda = "400 nm"" " < " " lamda = "600 nm"`

you can say that a photon with a wavelength of `"400 nm"` will be more energetic than a photon with a wavelength of `"600 nm"`.

To double-check this result, calculate the frequencies of the two photons

`nu = c/(lamda)`

Plug in your values to find -- do not forget to convert the wavelength to meters

`nu_"400 nm" = (3 * 10^8color(red)(cancel(color(black)("m")))"s"^(-1))/(400 * 10^(-9)color(red)(cancel(color(black)("m")))) = 7.5 * 10^(14)"s"^(-1)`

`nu_"600 nm" = (3 * 10^8color(red)(cancel(color(black)("m")))"s"^(-1))/(600 * 10^(-9)color(red)(cancel(color(black)("m")))) = 5.0 * 10^(14)"s"^(-1)`

Therefore, the energies of the two photons will be

`E_"400 nm" = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 7.5 * 10^(14)color(red)(cancel(color(black)("s"^(-1)))) = color(darkgreen)(ul(color(black)(5.0 * 10^(-19)"J"`

`E_"400 nm" = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 5.0 * 10^(14)color(red)(cancel(color(black)("s"^(-1)))) = color(darkgreen)(ul(color(black)(3.3 * 10^(-19)"J"`

I'll leave the answers rounded to two sig figs.

As you can see, the photon with the shortest wavelength had the highest frequency, and thus the highest energy.