Question #8d66e
1 Answer
Explanation:
For starters, you should know that the energy of a photon is directly proportional to its frequency as given by the Planck - Einstein relation
#color(blue)(ul(color(black)(E = h * nu)))#
Here
#E# is the energy of the photon#nu# is the frequency of the photon#h# is Planck's constant, equal to#6.626 * 10^(-34)"J s"#
Now, frequency and wavelength have an inverse relationship as given by the equation
#color(blue)(ul(color(black)(nu * lamda = c)))#
Here
#lamda# is the wavelength of the photon#nu# is the frequency of the photon#c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)#
This means that you can rewrite the Planck - Einstein relation using the wavelength of the photon
#nu * lamda = c implies nu = c/(lamda)#
to get
#E = h * c/(lamda)#
Plug in your value to find the energy of a single photon of wavelength
#4000 color(red)(cancel(color(black)("pm"))) * "1 m"/(10^12color(red)(cancel(color(black)("m")))) = 4 * 10^(-9)# #"m"#
and you will get
#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)))))/(4 * 10^(-9)color(red)(cancel(color(black)("m"))))#
#E = 4.97 * 10^(-17)# #"J"#
This means that in order to get a total of
#1 color(red)(cancel(color(black)("J"))) * "1 photon"/(4.97 * 10^(-17)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(2 * 10^(16)color(white)(.)"photons")))#
The answer is rounded to one significant figure, the number of sig figs you have for the wavelength of the photon.