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what i understood was the energy given off going between any two n levels is E = -(hR)/(n^2) so as you change n levels near infinity you don't really change and distance and when you don't move you don't do work so the energy change is 0. Honestly I'm not 100% clear on it either since I don't fully understand the negative, someone pls help us
I think it's because as n-->infinity, E becomes so small it approaches 0 (E-->0). When we look at the equation E=-(hR)/(n^2), we can see that as n were to increase and get so large (infinity), E would be equal to a fraction so small it would, in theory, approach 0. I think?
I think based off the equation E=-(hR)/n^2, E and n have an inverse relationship (As n increases, E will decrease. As n decreases, E will increase). So as n gets larger and larger and eventually approaches infinity, E will get smaller and smaller to the point where it is so minuscule, that it's approaching 0.
From a mathematical standpoint, the limit of the function as n approaches infinity is 0 because as the denominator of the function gets closer and closer to infinity, you get closer and closer to 0. From a conceptual standpoint I think energy approaches 0 as n goes to infinity because the energy levels get closer together as n goes to infinity which means less energy is absorbed/emitted.
When the electron is freed, its energy is at its maximum and its given as 0. And then for the energy level, the closer the electron is to the atom, the more it will be bound. So when the electron is free, the energy level can be said to be infinite because it's not bound to the atom at all.
In lecture, Lavelle mentioned once the energy of e- on the second energy level matches that of the first, the electron is excited. However, I was under the impression that one energy level had to be completely full before e- proceeded to the next energy level. Can someone explain what he meant by this?
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