Question #2e3c2

1 Answer
Dec 19, 2016

Here's what I got.

Explanation:

Actually, all you have to do here is convert the frequency of the photon to wavelength by using 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 #3.00 * 10^8"m s"^(-1)#

Keep in mind that you have

#"1 MHz" = 10^6"Hz" " "# and #" ""1 Hz" = "1 s"^(-1)#

which means that the frequency of the photon is equal to

#0.15 color(red)(cancel(color(black)("MHz"))) * (10^6color(red)(cancel(color(black)("Hz"))))/(1color(red)(cancel(color(black)("MHz")))) * "1 s"^(-1)/(1color(red)(cancel(color(black)("Hz")))) = 1.5 * 10^5"s"^(-1)#

Rearrange the equation to solve for #lamda#

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

Plug in your value to get

#lamda = (3.00 * 10^8 "m" color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^6color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(2.0 * 10^2"m")))#

The answer is rounded to two sig figs, the number of sig figs you have for the frequency of the photon.

Now, the second equation you listed

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

where

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

is called the Planck - Einstein relation and can be used to find the energy of a photon that has #nu# as its frequency.

You don't need to use this equation in order to find the wavelength of the photon, but you can rewrite it as a function of wavelength

#E = h * c/(lamda)#

This means that the wavelength of the photon can be written in terms of its wavelength as

#lamda = (h * c)/E#