Consider an electron undergoing cyclotron motion in a magnetic...

Question

Consider an electron undergoing cyclotron motion in a magnetic field. According to Bohr, the electron’s angular momentum must be quantized in units of ℏ. Find an expression for the allowed energy levels E_n in terms of ℏ and the cyclotron frequency f_cyc.

Accepted Answer

Hello, let's go through this practice problem according to bore the angular momentum of electrons exists in multiples of H bar. That is angular momentum is quantized assuming the angular momentum of protons exists in multiples of H bar. Derive an expression for the energy levels of the proton. When moving in circular paths around a magnetic field, A E sub N equals embnh bar divided by two B equal to emb divided by NH bar C equal to emb divided by two NH bar and Debnh bar divided by two M. So this is a somewhat tricky problem where we're trying to find a formula for the energy. But write it specifically in terms of these terms that involve each bar and angular momentum. So to start us off, let's consider the normal formula we use when the typical formula we use for the energy of a particle in motion. That is the formula for the particle's kinetic energy. And typically speaking, we represent kinetic energy, which would be just the same as the total energy of the particle as being equal to one half MV squared MV sub N squared. Now let's set this aside for now. And consider the forces acting on the particle. So it's moving around in a circle from the magnetic field. So let's consider the force that exists due to a magnetic field. And we'll specifically bring up Newton's second law, which states that the net force acting on a body is equal to the body's mass multiplied by its acceleration. In this case, that acceleration is a centripetal acceleration with a circular path. And centripetal acceleration is equal to the square of the speed divided by the radius of the distance the radius of that circular path. Now, in terms of the forces, this is just equal to one force, the magnetic force and recall that the magnetic force has a magnitude of the charge of the object. In this case, the elementary particle multiplied by the speed it's moving at V suban multiplied by the strength of the magnetic field. So now let's isolate this equation and do some canceling out. So one of these V subs NS will cancel out with one another. And let's also bring this R sub N into the right hand side of the equation. We do that. Then we find out something quite interesting. We end up with an equation saying that MV sub N is equal to EBR sub N. What's interesting about this is that our left hand side of the equation is just MV sub N. And according to something called the bore condition for the angular momentum of an electron MV. Sub N must be equal to N multiplied by H bar. So we substitute NH bar in for NV sub N in our EBR equation. And we find that EBR sub N squared is equal to NH bar. We're almost ready to substitute this equation into our energy equation and take the final answer. But in order to do with that, we first need to, we still have an R sub N and a variable that we're not supposed to use. So let's rewrite this equation to solve for R sub N so that we have something that we can just cancel out. So let's solve this for R sub N by dividing both sides of the equation by EB and then taking the square root of both sides. So dividing both sides, we find that our sub N squared is equal to NH bar divided by EB and then taking the square root, we can see that R sub N is equal to the square root of NH bar divided by E B. Now let's begin plugging some of this stuff into our energy equation. Recall that the energy equation we wrote earlier is one half MV sub N squared. Well, now we have an, an expression for the sub N into V sub N. We substitute E V Saban rather EB R sub N and that's all squared. And there's also a divided by M now not done yet, but real quick. Let's do some simplification. So now we have in the numerator me squared B squared R sub N squared divided by two M squared. So one of these MS will cancel out. And now let's plug in the formula that we'd found earlier for R sub M. So for our sub N, we found a bunch of things underneath a square root. But now it's being squared, let's just copy in NH bar divided by EB Yeah. EB so that would be E squared B squared divided by two M all multiplied by NH bar divided by EB. So one of these BS will cancel out and one of the es will cancel out. So what we're left with is Ebnh Bar divided by two M. So that is our answer. And if you look at the multiple choices, this agrees with option D. So option D is the correct answer to this problem. That is it for this video. I hope I helped you out and please check out our other tutoring videos for more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

Consider an electron undergoing cyclotron motion in a magnetic field. According to Bohr, the electron’s angular momentum must be quantized in units of ℏ. Find an expression for the allowed energy levels E_n in terms of ℏ and the cyclotron frequency f_cyc.

Key Concepts

Cyclotron Motion

Quantization of Angular Momentum

Energy Levels and Cyclotron Frequency

Watch next

Consider an electron undergoing cyclotron motion in a magnetic field. According to Bohr, the electron’s angular momentum must be quantized in units of ℏ. Find an expression for the allowed energy levels En in terms of ℏ and the cyclotron frequency fcyc.

Key Concepts

Cyclotron Motion

Quantization of Angular Momentum

Energy Levels and Cyclotron Frequency

Watch next

Consider an electron undergoing cyclotron motion in a magnetic field. According to Bohr, the electron’s angular momentum must be quantized in units of ℏ. Find an expression for the allowed energy levels E_n in terms of ℏ and the cyclotron frequency f_cyc.