Question

Question #3b89b

Answer 1

`7 * 10^(-20)"J"`

Explanation:

This is a great example of a practice problem for the Planck - Einstein relation, which establishes a relationship between the energy of a photon and its frequency.

More specifically, the energy of a photon, `E`, is said to have a direct relationship with the frequency of the photon, `nu`, as shown by the equation

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

Here

`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`

Simply put, the higher the frequency of the photon, the higher its energy.

In your case, the energy of a photon that has a frequency of `10^(14)"s"^(-1)` is equal to

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

`E = 6.626 * 10^(-20)"J"`

Now, the answer must be rounded to one sig fig, since that's how many sig figs you have for the frequency of the photon

`E = color(green)(|bar(ul(color(white)(a/a)color(black)(7 * 10^(-20)"J")color(white)(a/a)|)))`