Question #9af62
1 Answer
Here's what I got.
Explanation:
The thing to remember about frequency and wavelength is that they have an inverse relationship described by 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 * 10^(8)# #"m s"^(-1)#
In your case, the wavelength of the photons is given in nanometers, so start by converting it to meters
#545 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 5.45 * 10^(-7)color(white)(.)"m"#
Rearrange the equation to find the frequency of the photons
#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))/(5.45 * 10^(-7)color(red)(cancel(color(black)("m"))))#
#color(darkgreen)(ul(color(black)(nu = 5.50 * 10^(14)color(white)(.)"s"^(-1))))#
The answer is rounded to three sig figs, the number of sig figs you have for the wavelength of the photons.
Now, to find the energy of a photon of wavelength
#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"#
Plug in your values to find
#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 5.50 * 10^(14) color(red)(cancel(color(black)("s"^(-1))))#
#color(darkgreen)(ul(color(black)(E = 3.64 * 10^(-19)color(white)(.)"J")))#
Once again, the answer is rounded to three sig figs.