1
$\begingroup$

We know for a fact that if an electron absorbs a photon with enough energy it gets excited to a higher energy level, which increases that electron's total energy and we know that energy can present itself either with rest mass energy, kinetic energy and potential energy. And we know from Bohr equations that the higher the r(radius) is lower the linear velocity is,

so its kinetic energy should decrease and be converted into potential energy? If so , if it is just a matter of energy conversion between kinetic and potential energies how does the total energy of an excited electron increase?

How does the photon energy that excites an electron to a higher energy level shared between the potential and kinetic energy of that electron?

enter image description here

$\endgroup$
7
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Jun 2, 2023 at 10:40
  • $\begingroup$ @Community Edited. $\endgroup$ Commented Jun 2, 2023 at 10:50
  • $\begingroup$ Linked. $\endgroup$ Commented Jun 2, 2023 at 11:20
  • $\begingroup$ @CosmasZachos I am looking for a simpler explanation using Bohr Model. $\endgroup$ Commented Jun 2, 2023 at 11:22
  • $\begingroup$ The real QM situation should inform your Bohr model mental picture, as always. $\endgroup$ Commented Jun 2, 2023 at 17:35

1 Answer 1

0
$\begingroup$

"When an electron absorbs a photon and gets excited to higher energy level what does happen to its kinetic and potential energy?"

Here we talk about the total energy of the atom (because the potential energy term involves the nucleus as well). When you excite an atom with a photon, the atom's energy increases (because $E_{1}$ is negative - you can also say that the energy decreases in absolute value). $$E_{n}=\frac{E_{1}}{n^2}$$ You can also check what happens with the kinetic energy and potential energy individually using the expression for Bohr radius $r_{n}$ and the associated velocity $v_{n}$. $$r_{n}\propto n^2$$ If $r$ increases the potential energy increases (since the potential energy between electron and proton is negative).

$$U=-\frac{e^2}{4\pi\varepsilon_{0}r_{n}}$$

The velocity is of course lower for greater $n$, so its kinetic energy. But overall the total energy of the atom increases, because the potential energy increases more than the kinetic energy decreases.

$$v_{n}\propto \frac{1}{n}$$

It's not a matter of converting the kinetic energy to the potential energy of the same atom. It's about transfering energy from a photon to the atom.

$\endgroup$
5
  • $\begingroup$ But in the picture of the equation I provided in my OP says that velocity is directly proportional to n, not inversely proportional. $\endgroup$ Commented Jun 2, 2023 at 11:33
  • $\begingroup$ You might want to mention that the expressions for $r$, $U$, and $v$ are expected values. The atom is not in an eigenstate of any of those $\endgroup$
    – Dale
    Commented Jun 2, 2023 at 11:34
  • $\begingroup$ @medicalphysics, yes, but your equation contains $r$ in the denominator, which also depends on $n$, as you can see it in my answer. In the end, $v$ is inversely proportional to $n$. $\endgroup$ Commented Jun 2, 2023 at 11:38
  • $\begingroup$ @AWanderingMind Thank you. $\endgroup$ Commented Jun 2, 2023 at 11:40
  • $\begingroup$ @AWanderingMind You wrote, "But overall the total energy of the atom increases because the potential energy increases more than the kinetic energy decreases." Is this physically realistic? Does things like that disproportionate increase & decrease in KE and U happen in nature? Or is it a flaw that Bohr's theory has? Or is it about the Bohr theory being almost completely wrong? Can you elaborate on it a little, please? $\endgroup$ Commented Jun 3, 2023 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.