Question #819ac
1 Answer
Explanation:
Start by converting the wavelength of the photons to frequency by using the fact that wavelength and frequency have an inverse relationship described by 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)#
Rearrange to solve for
#nu * lamda = c implies nu = c/(lamda)#
Plug in your value to find -- do not forget to convert the wavelength from nanometers to meters
#nu = (3 * 10^8 color(red)(cancel(color(black)("m")))"s"^(-1))/(785 * 10^(-9)color(red)(cancel(color(black)("m")))) = 3.822 * 10^(14)# #"s"^(-1)#
Now, your next goal is to figure out the energy of a single photon of this frequency. The energy of a 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
Plug in your value to find
#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 3.822 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))#
#E = 2.532 * 10^(-19)# #"J"#
Finally, you know that the total energy emitted by the diode in the given time period is equal to
#31.0 color(red)(cancel(color(black)("J"))) * "1 photon"/(2.532 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(1.22 * 10^(20)color(white)(.)"photons")))#
The answer is rounded to three sig figs.