Atomic Structure and Electric Potential Energy

[pm1]

I get that,
Electrons have a negative charge and are attracted to the positive nucleus. Work is required to move an electron away from the nucleus, hence, the farther away an electron is positioned from the nucleus, the greater its potential energy.

however, when I was trying to make sense to this with the formula U=Kqq/r it seems to be telling me the opposite. If the electron is moved away from the nucleus isn't r getting bigger and consequently U getting smaller?

I thought about the negative sign of the electron. But, isn't U a scalar quantity? or is it vectorial? Then making an electron further from nucleus with a smaller *negative* number, hence bigger. 😕

Thank you!

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[QrtrLifeCrisis]

I get that,
Electrons have a negative charge and are attracted to the positive nucleus. Work is required to move an electron away from the nucleus, hence, the farther away an electron is positioned from the nucleus, the greater its potential energy.

however, when I was trying to make sense to this with the formula U=Kqq/r it seems to be telling me the opposite. If the electron is moved away from the nucleus isn't r getting bigger and consequently U getting smaller?

I thought about the negative sign of the electron. But, isn't U a scalar quantity? or is it vectorial? Then making an electron further from nucleus with a smaller *negative* number, hence bigger. 😕

Thank you!

I'll try to take this one, but I'm not exactly sure what your query is.

U=Kqq/r can also be stated as U=-Kqq/r. Both are used and depends on the convention of interest.

Taking U=-Kqq/r, which is better to use to explain this, as r increases, U goes to 0 from a negative number. Therefore, U is actually increasing to 0, which is the max U at r = infinity.

I wouldn't think about it in terms of scalar vs vector because that implies changes in direction when all this is is differences in convention.

 

[medemic]

I'll add my two cents for the sake of exercise. You are right about it being scalar, however, the negative sign denotes "attractiveness." I struggled with this concept myself. The electrical potential energy can describe two electrostatic relationships: repulsion and attraction. Repulsion is denoted by positive energy while attraction is denoted by negative (due to the negative sign). So, a highly negative electrical potential energy means strong attraction which makes sense if you place two oppositely charged particles close to each other. However, to pull the e- away, energy, in the form of work, must be put into the system to move the particles away from each other. As the post states above, as energy is put INTO the system, it comes closer to zero. Therefore, it becomes more positive in a sense, and in this regard, you can say its gaining potential energy. Note that the energy will still be negative. Why? For oppositely charged particles, the maximum potential energy a negatively charged particle can have is 0 electrical potential energy. 0 electrical potential energy means that the electron (in this case) is no longer in its "bound" state. Oppositely charged particles are either bound (attracted) or unbound (completely ionized). So going to to the equation, as r increases, the U is becoming less negative, which means its gaining energy (from the work out into moving it far away from the attractive positively charged particle). I hope this makes sense. Let me know which part you don't understand. This concept has a lot of application.