Question #41bbe
1 Answer
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
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.
