Question

Question #429d3

Answer 1

`E = 2.3 * 10^(-8)"J"`

Explanation:

The first you need to do here is to figure out the energy of a single photon that travels at that particular wavelength.

To do that, use the Planck - Einstein equation, which looks like this

`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

Now, the problem provides you with the wavelength of the photon, which means that you're going to convert this to frequency by using the fact that wavelength and frequency have an inverse relationship as given by

`color(blue)(ul(color(black)(lamda * nu c)))`

Here

• `lamda` is the wavelength of the photon
• `c` is the speed of light in a vacuum, usually given as `3.0 * 10^8"m s"^(-1)`

In your case, a wavelength of

`1870 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9 color(red)(cancel(color(black)("nm")))) = 1.870 * 10^(-6)"m"`

will correspond to a frequency of

`nu = c/(lamda)`

`nu = (3.0 * 10^8 color(red)(cancel(color(black)("m")))"s"^(-1))/(1.870 * 10^(-6)color(red)(cancel(color(black)("m")))) = 1.604 * 10^(14)"s"^(-1)`

This means that the energy of a single photon will be equal to

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 1.604 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))`

`E = 1.063 * 10^(-19)"J"`

Use the total number of incoming photons to calculate the total energy detected per second

`6. * 10^(7) color(red)(cancel(color(black)("photons"))) * (1.063 * 10^(-19)"J")/(1color(red)(cancel(color(black)("photon")))) = 6.378 * 10^(-12)"J"`

This means that in

`1 color(red)(cancel(color(black)("h"))) * (60 color(red)(cancel(color(black)("min"))))/(1color(red)(cancel(color(black)("h")))) * "60 s"/(1color(red)(cancel(color(black)("min")))) = "3600 s"`

the total energy detected by the device will be

`3600 color(red)(cancel(color(black)("s"))) * (6.378 * 10^(-12)"J")/(1color(red)(cancel(color(black)("s")))) = color(darkgreen)(ul(color(black)(2.3 * 10^(-8)"J")))`

The answer is rounded to two sig figs.