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Emily Ng_4C
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am


Postby Emily Ng_4C » Tue Oct 16, 2018 10:12 am

Why is there a negative in E=-hR/n^2?

JT Wechsler 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

Re: Formula

Postby JT Wechsler 2B » Tue Oct 16, 2018 10:16 am

When an electron is still around the nucleus, the system is more stable and at a lower energy than if it were to not be there at all. Because of this, there is a negative sign in the equation.

Kelsey Warren 1I
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

Re: Formula

Postby Kelsey Warren 1I » Tue Oct 16, 2018 1:21 pm

The equation is (usually?) used for an electron losing energy and consequently emitting a photon. Because the electron loses energy, its energy is "negative" while the energy for the photon is positive. Dr. Lavelle also said in lecture that the negative sign means that the bound electron has lower energy than the free electron, in the case of an electron popping off the atom. I could use some more clarification on that as well.

Hailey Boehm 2H
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

Re: Formula

Postby Hailey Boehm 2H » Tue Oct 16, 2018 2:46 pm

The negative sign is there because a bound electron has lower energy than a free electron. When e- is completely removed, E=0. This equation makes 0 a reference point, and indicates that when the e- gets closer to nucleus/lower in orbitals, the e-'s energy is becoming more negative (or in other words it is losing energy).

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