Question #1e68a
1 Answer
Explanation:
The thing to remember here is that the energy of a photon is directly proportional to its frequency and, consequently, inversely proportional to its wavelength.
This relation is described by the Planck - Einstein relation, which looks like this
#E =h * c/(lamda)#
Here
#E# is the energy of the photon#h# is Planck's constant, equal to#6.626 * 10^(-34)"J s"# #c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)# #lamda# is the wavelength of the wave
Now, notice that the speed of light in a vacuum is expressed in meters per second. This tells you that you need to convert the wavelength of the photon from nanometers to meters by using the fact that
#"1 m" = 10^9color(white)(.)"nm"#
Plug in the value you have for the wavelength of the photon and solve for
#E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(500 * 10^(-9)color(red)(cancel(color(black)("m"))))#
#E = color(darkgreen)(ul(color(black)(4 * 10^(-19)color(white)(.)"J"#
The answer is rounded to one significant figure, the number of sig figs you have for the wavelength of the photon.