Question

Question #f8475

Answer 1

`5 * 10^(17)`

Explanation:

The idea here is that you need to use the wavelength of the photon to find the energy of a single photon, then use the total amount of energy required to figure out how many photons are needed.

You know that the energy of a photon is directly proportional to its frequency, which implies that it's inversely proportional to its wavelength, since

`color(blue)(ul(color(black)(nu = c/(lamda))))`

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)`

The Planck - Einstein relation allows you to calculate the energy of the photon using its wavelength

`color(blue)(ul(color(black)(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"`

Plug in the value you have for the wavelength of the photon

`100 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9 color(red)(cancel(color(black)("nm")))) = 1 * 10^(-7)` `"m"`

to find the energy of a single photon of this wavelength

`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)))))/(1 * 10^(-7)color(red)(cancel(color(black)("m"))))`

`E = 1.99 * 10^(-18)` `"J"`

You can now use the energy of a single photon to find the number of photons that would generate `"1 J"` of energy.

`1 color(red)(cancel(color(black)("J"))) * "1 photon"/(1.99 * 10^(-18)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(5 * 10^17color(white)(.)"photons")))`

The answer is rounded to one significant figure, the number of sig figs you have for your values.

As an interesting note, you can compare this result to the answer given here.

Notice that the wavelength of the photons is longer here

`"100 nm " > " 4000 pm = 4 nm"`

which is equivalent to saying that these photons have a lower frequency, so you need more photons

`5 * 10^17color(white)(.)"photons"" " > " "2 * 10^16color(white)(.)"photons"`

to generate `"1 J"` of energy.