Question #b058f

1 Answer
May 30, 2017

#1.32 * 10^(-6)"m"#

Explanation:

Start by calculating the energy of a single photon in joules.

You know that if you get #1# mole of photons at this wavelength, you get

#90.5 color(red)(cancel(color(black)("kJ"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) = 9.05 * 10^4# #"J"#

of energy. As you know, #1# mole of photons will contain #6.022 * 10^(23)# photons, as given by Avogadro's constant. This implies that the energy of a single photon will be equal to

#1 color(red)(cancel(color(black)("photon"))) * (9.05 * 10^4color(white)(.)"J")/(6.022 * 10^(23)color(red)(cancel(color(black)("photons")))) = 1.503 * 10^(-19)# #"J"#

Now, the energy of a photon is directly proportional to its frequency, as described by the Planck - Einstein equation

#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"#
  • #nu# is the frequency of the photon

Rearrange to solve for the frequency of the photon

#E = h * nu implies nu = E/h#

Plug in your values to find

#nu = (1.503 * 10^(-19) color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J"))) "s") = 2.268 * 10^14# #"s"^(-1)#

Finally, to find the wavelength of the photon, use the fact that frequency and wavelength have an inverse relationship that can be described by the equation

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

Here

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

Rearrange to solve for the wavelength of the photon

#nu * lamda = c implies lamda = c/nu#

Plug in your values to find

#lamda = (3 * 10^8 color(white)(.)"m" color(red)(cancel(color(black)("s"^(-1)))))/(2.268 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(1.32 * 10^(-6)color(white)(.)"m")))#

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