Energy of and electron is $$E = \text{kinetic energy} + \text{negative of potential energy}. \tag{eq-1}$$ But energy of electron in the $n$th orbital is also $$E = -\frac{13.6}{ n^2} \tag{eq-2}$$ Which means that the energy of the electron would get less and less negative as we go to further orbitals, which means that the kinetic energy of the electron is increasing (from eq-1). However that is counter-intuitive because as the electron goes from lower orbital to higher its going from a lower potential (closer to the nucleus) to a higher potential (away from nucleus i,e. against the electric field). And thus the potential energy should increase and the total energy should be more negative.

Another thing which is mind boggling is that the velocity of electron in the $n$th orbit is, $$v = \frac{e^2}{nh\epsilon_0}$$ which suggests that velocity decreases as we go from inner to outer orbitals, which means that kinetic energy decreases! All this does not make sense to me. I know I made a mistake somewhere so if someone can help me please explain.

At the same time my textbook says that the outer energy level electrons would have more energy so, they would need less energy to be excited. But what energy do they have? The velocity equation suggests that outer electrons have less and less kinetic energy as we go to oter orbitals and eq 2 suggest that the electron even decrease in potential energy as they go in further energy levels(which should increase)


1 Answer 1


You write

E = kinetic energy + negative of potential energy. (eq-1)

and then

as the electron goes from lower orbital to higher […] the potential energy should increase and the total energy should be more negative.

Remember that the electrostatic potential energy between two charges $q_1,q_2$ separated by some distance $r$ is

$$ U = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r} $$

This potential is not strictly positive: it has the same sign as the product of charges $q_1\cdot q_2$. From the definition of the force $\vec F = -\vec\nabla U$ (or, in one dimension, $\vec F = -\hat x\frac{\mathrm d}{\mathrm dx}U$), you should be able to convince yourself that this sign convention is required to make opposite charges attract and like charges repel.

When an electron moves from a lower to a higher orbit, the potential energy increases by becoming less negative. Your equation (eq-1) is wrong. The total energy is kinetic plus potential; it’s just that the potential energy between opposite charges happens to be negative.

  • $\begingroup$ "A negative potential energy means that work must be done against the electric field in moving the charges apart!" I quote is from a website. And this fits perfectly with the example of atom, work is being done to keep the electron against the electric field and collapsing into the nucleus, so our potential energy is negative but as we move to a further orbital more work is done to keep the electron against the electric field of the nucleus, so it should be more negative. $\endgroup$ May 16, 2022 at 13:55
  • $\begingroup$ youtu.be/qveyVsOahI8 This is for gravitational potential but it applies to charges too i guess $\endgroup$ May 16, 2022 at 14:00
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    $\begingroup$ No, the CHANGE in PE is negative when the work of the field is positive and the CHANGE is positive when the work of the field is negative. The actual sign of the PE depends on the reference and is completely arbitrary, in general. For the electrostatic PE with two charges of opposite signs and reference at infinite distance, the PE is negative everywhere and it increases as you increase the distance. It does not change sign at all. $\endgroup$
    – nasu
    May 16, 2022 at 14:36
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    $\begingroup$ The work done by a force is $W=\int \vec F\cdot \mathrm d\vec x$. If you are careful with signs and caveats about conservative forces, this is equivalent to the relationship between force and potential energy given in the answer. What matters is the sign of the change in the potential energy. You can change the sign of the energy itself by adding a constant offset. For instance, near Earth’s surface we may choose the gravitational “zero” to be the floor or the tabletop instead of $r\to\infty$ from Earth’s center. $\endgroup$
    – rob
    May 16, 2022 at 14:37
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    $\begingroup$ Yes, the PE decrease when the charges get closer and increases when they get further away. You can get more KE if you release one charge when at a larger distance from the other one. The maximum energy is at infinite distance and this maximum is zero. $\endgroup$
    – nasu
    May 16, 2022 at 14:47

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