Question #3b89b
1 Answer
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,
#color(blue)(|bar(ul(color(white)(a/a)E = h * nucolor(white)(a/a)|)))#
Here
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
#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)|)))#