The valence electron in potassium requires 6.95 xx 10^-19 "J" of energy to be removed. What frequency of light can do this? What type of electromagnetic radiation could do this?

1 Answer
Jul 3, 2017

Very blue visible photons or UV photons.

Explanation:

K electron configuration: [Ar]4s^1

This electron in the 4s orbital requires E = 6.95*10^-19 J to be emitted. This is merely a question of the photon's energy being that which meets this requirement, thus,

6.95*10^-19 J = (6.626*10^-34 J*s)v
v = 1.05*10^15 s^-1

Now, referring to the EMR diagram that I gaurantee you most people don't have absolutely memorized:
WikipediaWikipedia

This energy corresponds with the frequency of very blue photons (visible) or UV photons.