Good morning According to the Bohr model of the hydrogen atom, there is one electron in
orbit around one proton and the proton is the nucleus of the atom.
We are going to determine the speed at which the electron moves.
Bobby, please read the problem and Billy, please translate.
♪flipping physics♪
Assuming a circular orbit of the electron about the nuclear proton in the Bohr model
of the hydrogen atom, determine the speed of the electron.
The electron orbits at a radius of 5.29 × 10^–11 m.
- Okay. The radius of the orbit of the electron equals 5.29 × 10^–11 m, and we are looking for
the speed of the electron but it is moving in a circle.
So, the magnitude of the tangential velocity of the electron equals question mark.
- Aren't electrons in like clouds or something around the nucleus of the atom and not really orbits?
- Bo, with regards to the electron clouds, they describe how we do not actually know
the positions of the electrons in atoms.
However, we know probabilities for where the electrons might be.
Basically, we can make educated guesses for electron locations.
The Bohr model of the atom with electrons in orbital shells is from classical physics
and is an older less accurate model of an atom.
Older than the newer more accurate electron clouds from quantum physics.
- Wait a second, if the Bohr classical physics model is older and less accurate than the
electron clouds from quantum physics, why would we use it?
- Bo, that is a fair question.
So, we still use projectile motion equations even though we know air resistance exists.
We still model many pendulums as simple harmonic motion even though simple harmonic motion
is an approximation.
We model objects by discussing what happens to their center of mass rather than the whole object.
Often, we use basic models of the world to help us understand what is happening even
if those models are not perfect.
In other words, we know electrons in orbital shells may not quite be correct, however we
can still learn a lot from modeling the hydrogen atom that way.
- Okay.
- Okay. Bo, please solve for the speed of the electron.
- I don't really know where to start.
- The electron is moving in a circle around the proton.
Maybe, you start with a free body diagram of the forces acting on the electron and then
sum the forces on the electron in the indirection.
- It's a good idea, Bobby. Thanks.
- You are welcome.
- The only force acting on the electron is the electric force which is directed in toward the proton.
So, the net force in the indirection on the electron equals the electric force.
And the net force in the indirection also equals the mass of the electron times acceleration
in the indirection which is the centripetal acceleration.
For the electric force, we can substitute the Coulomb constant times charge one times
charge two divided by the square of r, the distance between the centres of charge of
the two charges, the proton and the electron.
And for the centripetal acceleration, we can substitute in
tangential velocity squared divided by radius.
Oh... In this instance, r, the distance between the centres of charge of the electron and the
proton is also the radius of the orbit of the electron.
So this is an example where r is also a radius. Nice.
Oh, and....
- Everybody brought one of our r to the party.
Everybody brought mass...
- Great. Billy, take it from here please.
- Oh, good. Now we can talk - - Hold on Billy.
What about the force of gravity which is pulling the proton and electron towards one another?
- And the force of gravity pulling the electron and proton down towards the center of the earth?
- Remember when we showed the Coulomb constant was way larger than the universal gravitational constant?
I think the point there was that the electric force is so much larger than those gravitational
forces that we can ignore them.
Right Mr. P?
- That is correct Billy.
When it comes to electric forces between atomic particles, often gravitational forces are
so small they are negligible. Billy?
- Okay. We can solve for the tangential velocity of the electron by dividing the equation by mass
and taking the square root of the whole equation.
We get tangential velocity equals the square root of the Coulomb constant times charge
one times charge two divided by the quantity mass times radius.
With numbers that is the square root of the quantity, 8.99 × 10^9, times... well, both
of those are the magnitudes of the charges of the proton and electron which are both
1.6 × 10^-19 C. And so, we square that number.
- But the charge of a proton is positive and the charge of an electron is negative.
So, one charge should be negative and one should be positive, right?
- Actually, when I drew the freebody diagram and summed the forces in the indirection,
I already defined the direction of the electric force as inward and therefore a positive.
Therefore, because we have already defined the direction of the electric force in the
equation, we use the magnitudes of those charges.
- Sure that makes sense. Thanks, Bo.
- You are welcome.
- Okay. We divide by the quantity mass of an electron because that is what we summed the forces
on times the radius of the orbit of the electron.
However, I do not think the mass of an electron was given in the problem.
- The Coulomb constant and elementary charge were also not given in the problem.
I think he assumes we can look up standard numbers like these.
- Actually, if you are taking an AP exam, constants like these are in the constants
and conversion factors section of the equation sheet which you will always have on the AP exam.
- Look who has been studying...
- That is very helpful Bo.
- Yap.
- Okay. So, the mass of an electron is 9.11 × 10^-31 kg and the radius of the orbit of the electron
was given in the problem as 5.29 × 10^-11 m.
That means, the speed of the electron equals 2.1853 × 10^6 or 2.19 × 10^6 m per second
with two significant digits.
- Wow! That is fast.
- Almost five million miles per hour.
So, yeah, that's fast.
- Yeah. That is fast.
- That's it. Thank you all for your hard work.
And thank you very much for learning with me today.
I enjoyed learning with you.