Question #80df2

1 Answer
Apr 9, 2017

Here's what I got.

Explanation:

For starters, you should know that wavelength and frequency have an inverse relationship that can be described using 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)#

This means that photons that have a higher frequency will have a shorter wavelength and photons that have a lower frequency will have a longer wavelength.

Notice that the speed of light is expressed in meters per second, so make sure to convert the wavelength from nanometers to meters before using the equation.

#710 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9 color(red)(cancel(color(black)("nm")))) = 7.10 * 10^(-7)# #"m"#

Rearrange the equation to solve for #nu#, the frequency of the photon

#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))/(7.10 * 10^(-7)color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)(4.2 * 10^(14) color(white)(.)"s"^(-1))))#

Now, the energy of the 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

This time, you can say that photons that have a higher frequency will have a higher energy than photons that have a lower frequency.

Plug in your value to find

#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 4.2 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))#

#E = color(darkgreen)(ul(color(black)(2.8 * 10^(-19)color(white)(.)"J")))#

Both values are rounded to two sig figs, the number of significant figures you have for the wavelength of the photon.