Question

Question #819ac

Answer 1

`1.22 * 10^(20)`

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`

`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 J"`, which means that the number of photons needed to get this energy output 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.