Question

Question #694db

Answer 1

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

Explanation:

The relationship between the energy of a photon and its frequency is given by 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

As you can see, the energy of the photon is directly proportional to its frequency. This means that photons that have higher frequencies will also be more energetic than photons that have lower frequencies.

All you have to do here is plug in the value you have for the frequency of the photon and solve for `E`

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 6.00 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))`

`color(green)(bar(ul(|color(white)(a/a)color(black)(E = 3.98 * 10^(-19)"J")color(white)(a/a)|)))`

The answer is rounded to three sig figs.

Notice that the energy of the photon is expressed in joules, `"J"`, an SI derived unit of energy. This comes about because the unit of seconds, `"s"`, used in the expression of the Planck constant and the unit of 1 over second, `"s"^(-1)`, used for frequency cancel each other.