Question

Question #f8bdb

Answer 1

`2.4 * 10^5"J"`

Explanation:

The first thing to do here is to figure out the energy of a single photon by using the Planck - Einstein relation

`color(blue)(bar(ul(|color(white)(a/a)color(black)(E = h * nu)color(white)(a/a)|)))`

Here

`E` - the energy of the photon
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`
`nu` - the frequency of the photon

You know that the photon has a frequency of `6 * 10^(14)"s"^(-1)`, so plug in this value to find its energy

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 6 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))`

`E = 3.98 * 10^(-19)"J"`

Now, one mole of photons consist of exactly `6.022 * 10^(23)` photons, as given by Avogadro's constant.

This means that the energy of one mole of photons will be

`6.022 * 10^(23) color(red)(cancel(color(black)("photons"))) * (3.98 * 10^(-19)"J")/(1color(red)(cancel(color(black)("photon")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.4 * 10^(5)"J")color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the frequency of a photon.