Why does larger n give lower energy in the hydrogen atom equation?

[Jules18]

Here's the equation I'm dealing with (it describes the energies that an electron in a hydrogen atom can occupy) :

E_n = -R_H(1/n²)

The way I understood, the bigger n was the farther away the e^- was from the nucleus, so it would have more potential energy.
But n is in the denominator in this eq'n, so the bigger the n the less energy the electron occupies.

So I'm kinda confused. Help?

~Jules~

 

[Nabeshin]

Here's the equation I'm dealing with (it describes the energies that an electron in a hydrogen atom can occupy) :

E_n = -R_H(1/n²)

The way I understood, the bigger n was the farther away the e^- was from the nucleus, so it would have more potential energy.
But n is in the denominator in this eq'n, so the bigger the n the less energy the electron occupies.

So I'm kinda confused. Help?

~Jules~

Note the negative sign in the equation. Potential is zero at r=infinity, and indeed as r increases, so does energy.

 

[Jules18]

okay. I'll take your word for it.

 

[jtbell]

Why take his word for it? Plug in some numbers and see what energies you get for n = 1 and n = 2!

 

[Jules18]

yeah, I did a little while ago and the negative sign still bugs me a little bit.

But I'm sort of able to grasp why it's there ... it's to make sure the difference will be positive when your final n is higher than your initial n, right?

or is it to make sure that even thought the absolute value of the energy at a lower n is higher, the negative sign makes it technically lower?

 

[Pengwuino]

The electron is bound to the system, meaning it has negative energy. If it were able to get an infinite distance away, which corrosponds to n -> infinity, it would have 0 energy. For example, imagine a planet. If an object orbits that planet, it has a negative energy because it's bound (all this of course, taking V = 0 at infinity). If it had some kinetic energy greater then the magnitude of the potential energy, it would obviously be free from the orbit and no longer bound. Now, of course, by no means is this the same situation, I'm just hoping to clarify why it's negative and what happens as n-> infinity.

 

[vanesch]

yeah, I did a little while ago and the negative sign still bugs me a little bit.

But I'm sort of able to grasp why it's there ... it's to make sure the difference will be positive when your final n is higher than your initial n, right?

or is it to make sure that even thought the absolute value of the energy at a lower n is higher, the negative sign makes it technically lower?

Mmm, seems you're somewhat confused about negative numbers ?

What's the highest value, -5 or -2 ? Do you INCREASE or DECREASE X when X goes from -5 to -2 ?

 

[Jules18]

increase?

 

[vanesch]

increase?

Right

So when you go from -1/4 to -1/9, you also increase, right ?

 

[jtbell]

As a check, plot them along the vertical axis of a graph. Which one is "higher"?