INT In a classical model of the hydrogen atom, the electron or...

Question

INT In a classical model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius 0.053 nm. What is the orbital frequency in rev/s? The proton is so much more massive than the electron that you can assume the proton is at rest.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a nanoscale electrostatic system. A tiny charged particle similar to an electron orbits, a central charged core similar to a proton in a circular orbit with a radius of 0.053 nanometers. The central core's mass is significantly greater allowing it to be considered at rest. What is the orbital frequency of the charged particle in revolutions per second? OK. So that is our end goal is to find the orbital frequency of the charged particle in revolutions per second. OK. So we're given some multiple choice answers here. They're all given in the same units of Hertz. Let's read them off to see what our final answer might be. A is 4.45 multiplied by 10 to the power of 15 B is 6.15 multiplied by 10 to the power of 17 C is 7.51 multiplied by 10 to the power of 15 and D is 6. multiplied by 10 to the power of 15. OK. So first off, let us recall and use Kums Law equation considering the attraction between an electron and proton in a hydrogen atom. And let's call it equation one. And that states that the force is equal to K where K is Kums constant multiplied by the absolute value of the charge of a proton multiplied by the charge of an electron. OK. All divided by the radius of the orbit squared. OK. So we also need to recall and use the equation for the Syed force or the electron circular motion. And let's call that equation two. So the syrup force states that the force is equal to the mass of an electron multiplied by the angular frequency multiplied by the radius of the orbit. OK. So now we need to set equations one and two equal to each other. So that will look like this would be plum's constant multiplied by the absolute value of charge of the proton multiplied by the charge of the electron divided by yeah radius of the orbit squared equal to the mass of an electron multiplied by the angular frequency squared multiplied by the radius of the orbit. But we need to make a quick note here that the angular frequency is equal to two pi multiplied by the frequency. So let's plug in two pi multiplied by the frequency in for the angular frequency. And let's call that equation three when we make that change Kums constant multiplied by the absolute value of charge of the proton multiplied by the charge of the electron divided by the radius of the orbit squared equal to the mass of the electron multiplied by the angular frequency which is two pi F squared multiplied by the radius of the orbit. Now we need to rearrange equation three to solve for frequency. So when we do that to get frequency all by itself using a little bit of algebra, we get that F is equal to the square root of Kums constant multiplied by the absolute value of the charge of a proton multiplied by the charge of an electron divided by or pi squared multiplied by the mass of an electron multiplied by the radius of the orbit tubed. OK. OK. So now we can plug in our known variables and solve for F. So let's do that. So F equals, and let's recall that Plum's constant is 9.0 multiplied by 10 to the power of nine newtons multiplied by meter squared. The per Coolum squared multiplied by. And let's note here since this is the absolute value of the charge of a proton multiplied by the charge of an electron. But since we can use K, all right. So correction. So we need to note here since it's the charge of the proton and the electron are similar, we can just state that and it's 1.60 multiplied by 10 to the power of nine columns squared. So we'll just use the charge of a proton and just square it because it says that the charge of a of the particles similar to electron. So we're actually gonna use the charge of an electron and just square it divided by four pi squared multiplied by the mass of an electron which is 9.11 multiplied by 10 to the power of negative kg multiplied by the radius of the orbit which is given to us in nanometers. So we need to convert it to meters. So we need to take 0.053 and multiply it by 10 to the power of negative nine, convert it to meters. So when we take the square root of that and type it into a calculator, we should get that. The frequency is 6.55 multiplied by 10 to the power of 15 Hertz. And this is the same as revolutions per second. So that is our final answer. So when we go back up to look at our multiple choice answers, that means that our final answer has to be the letter D 6.55 multiplied by 10 to the power of 15 Hertz. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

INT In a classical model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius 0.053 nm. What is the orbital frequency in rev/s? The proton is so much more massive than the electron that you can assume the proton is at rest.

Key Concepts

Orbital Frequency

Centripetal Force

Bohr Model of the Atom

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