2026 AP Physics 2 Free Response #2

Hinzugefügt: 2026-05-11

[00:00:00]All right, we're doing the second FRQ of the AP Physics 2 2026 version J of the exam. That's the one that is released.

[00:00:07]None of the other versions are released.

[00:00:09]So, figure one represents energy levels corresponding to state for a hypothetical atom. Draw arrows to represent all possible atomic transitions that could result in the emission of a photon. Emission of a photon means we are going to lose energy cuz the photon has energy. So, we're going to drop energy states. So, we're going to go from here. We could also go from here. We could also be at n = 2 and go to n = 1. So, any possible way we drop, it's going to be one or you can just draw them here. I'll just draw it.

[00:00:35]Maybe I shouldn't draw it next to it. I should draw on the diagram. So, we'll just draw it. One, two, and three. Those are the three possibilities of how you could emit a photon, okay?

[00:00:46]Derive an expression for the wavelength of the highest energy photon that can be emitted from the atom. Express your answer in terms of E0 physical constants as appropriate. So, the energy is going to be um HF for a photon, which because we want it in terms of the wavelength, um we know that the speed of light the or for waves velocity is wavelength times frequency. So, speed of light, which is how fast a photon goes, speed of light, is lambda times F. And so, then we know that the way the frequency we can get rid of the frequency is C over lambda and plug that into here. So, it's going to be HC over lambda. You could have just gone straight into there, but I always think of the energy of the photon in terms of frequency. So, I always have to do that translation a little bit in there.

[00:01:28]Now, what is the biggest energy drop we got here? With minus 2 E0 and minus 5 E0, that's a change of 3 E0. So, the amount of energy is 3 E0 in that greatest energy photon is going to be HC over lambda. And then you just bring the lambda up and bring the 3 E0 down. You get HC over 3 E0, okay? Like that.

[00:01:49]Following parts applies to C and D.

[00:01:51]Device can emit a monochromatic electromagnetic radiation. The wavelength lambda of the radiation can be varied with a continuous range of wavelengths.

[00:01:59]On figure two, sketch the energy of the emitted photons from the device as a function of lambda for lambda to four lambda naught. Photons with wavelength lambda has four E0.

[00:02:08]Um so this is just making sure you understand this relationship, right?

[00:02:11]This HC over lambda is that the greater the wavelength, the less energy it has. So when it's lambda zero, it has Wait, hang on.

[00:02:20]Uh oh yeah, five E0.

[00:02:24]Um continuous wavelengths.

[00:02:28]So we have energy drops.

[00:02:31]Take a look at what what they're saying here.

[00:02:34]Okay, so lambda naught is four E0. I I was trying to see if it was related to the the first one at all. I don't think it has anything to do with it. You just want to know it's like one over. So when you go to four four lambda naught, because you you quadruple it, it's going to be four times as much. If you double it, it should be half as much. If you three times, it should be four-thirds, which is probably around like four and a third. I don't know. So it should have like a curve that looks like this.

[00:03:00]Okay.

[00:03:02]Um yeah, student claims that based on the energy state shown in figure one, the atom can emit a photon of wavelength lambda naught energy four E0 like the device described in part C. Indicate whether the claim is correct or incorrect.

[00:03:12]Um Oh, they are relating you to part A, cuz they want you to include what claim representation in part A.

[00:03:23]Uh the atom. Okay, so now we're referring back to the atom. Okay, so they're saying can the atom follow this this diagram here?

[00:03:30]Well, the atom can only do three E0, two E0, or one E0. It does not have the ability to generate a photon with four E0, cuz there's no jump that's going to give you four E0 of energy change, right? And so in that case, no, it's not possible. So it is incorrect.

[00:03:51]Um the only energy um state changes in part A are 3 E0, 2 E0, and E0.

[00:04:11]There is no way for Uh, they said briefly, so you don't have to write that much. There would be to it to emit a photon of 4 E0.

[00:04:28]Okay, and that's it.