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# Bohr model radii (derivation using physics)

Learn about the Bohr model of the hydrogen atom and the physics behind it. Use equations such as Coulomb's law and Newton's second law, along with the assumption that angular momentum of an electron is quantized, to calculate the radius of the atom for a ground state electron. Created by Jay.

## Want to join the conversation?

• At , "Bohr thought this angular momentum should be quantized." Why does he get to do that? How does one know that "quantizing" a parameter is appropriate, and how does one know HOW to quantize it?
(103 votes)
• it comes from quantum physics. This is also done with momentum, not only angular momentum
(2 votes)
• why coulomb's law is really similar to the law that describes the gravitation between two object F=G*(m1*m2)/r^2
(22 votes)
• Nice observation - it simply is. Charge and graviational attraction are two very similiar concepts!
(18 votes)
• When we multiplies e by negative e we should have got -e squared right? Where did that negative sign go?
(8 votes)
• If you are referring to the calculation at time , he pointed out that the interest is in the magnitude of the charge; hence, he took the absolute values (as indicated by the parallel vertical lines on the left side of the equation).
(23 votes)
• What is n in the equation?? I did not quite get it
(12 votes)
• Bohr and the science community at the time already knew that energy from the H atom was emitted at specific, discrete values (referred to as the wavelength or frequency of the emitted light energy). He and his mentor, Rutherford, had already conceived of the "orbital model" of atoms (as opposed to the earlier plum pudding model where all electrons and and protons are mixed together in the nucleus like raisins in a pudding). So, based on these he took an insightful hypothesis that the orbits that the electrons were in were orbits that were in incremental, or quantum if you prefer, steps. So his hypothesis involved asserting that the electrons were in an orbital distance of 1 radius, or 2 radii or 3 radii from the nucleus. It was an ingenious insight, but a hypothesis is an educated guess based on the observations. He needed to express it mathematically and compare that mathematical result with the observations of the energy emissions.

Like many hypotheses, it worked very well to describe the phenomena in a set of instances, but it failed to explain the in the broader and more general cases. It still provides a good basis for understanding effects, but in more complex atoms it did not work out. Rather than take a Ptolemy approach and add more and more conditions onto the hypothesis to explain the more complicated cases, the Bohr model was replaced with a fuller quantum explanation, in a sense like Copernicus replace the Ptolemaic hypotheses of a geocentric system.
(10 votes)
• How did Bohr arrive at the conclusion that L(angular momentum)=mvr=nh/2pi? I tried hard to make sense but still didn't get it. Please help
(10 votes)
• L is defined to be r x p, which is r*p*sin(theta), where theta is the angle between the radius vector and the momentum vector. Since they are moving in a circle, that means that p and r are perpendicular, so sin(theta) is just 1, leaving rp. Since p is just mv, that means that L=mvr.

The way he arrived at the conclusion L=nh/2pi is experimentally. The predicted emission spectrum of Hydrogen would not agree with reality unless he quantized L like that.
(11 votes)
• does first orbit of all elementns has same radius as in hydrogen?
(6 votes)
• Nice question

I dont think so. Each time to add an electron, the energy level of the system changes and, as far as I recall, the orbits get smaller.
(8 votes)
• What is angular momentum ?
(3 votes)
• It is the equivalent of momentum (mass*velocity) but because velocity is in one direction, and circular movement changes direction continuously (going in circles, not in a straight line) and therefore you use angular momentum instead.
(4 votes)
• how was the bohr's angular momentum quantization derived??
(1 vote)
• Bohr guessed that angular momentum might be quantized, and then he calculated the consequences of that in order to see if they matched up with experimental observations. His guess did a pretty good job of explaining the emission spectrum of hydrogen atoms.
(7 votes)
• how Bohr came up with L=n*h/2*pi. Did he discover this equation?
(3 votes)
• He guessed that it must be true because if it were true, and if the electrons traveled in little orbits around the nucleus, then that would explain why the emission spectrum is discrete. It turns out he was wrong about the orbits but his overall idea that the momentum of the electron has to be quantized was right.
(3 votes)
• I think you may have made a mistake. A normal velocity is equal to displacement multiplied by time. In the video, you said that the quantized velocity is equal to (n*h)/(2*pi*m*r), but I noticed that it doesn't involve time, just things that relate to displacement and, oddly, mass. How is this velocity?
(2 votes)
• Look at the units involved in Planck's constant.
(5 votes)

## Video transcript

- In the Bohr model of the hydrogen atom we have one proton in the nucleus. So I draw in a positive charge here and a negatively charged electron orbiting the nucleus, so kind of like the planets orbiting the sun. Even though the Bohr model is not reality it is useful for a concept of the atom. It's useful to calculate, say for example we can calculate the radius of this circle. We're actually gonna do that in this video. It's worth going in to some of the details. But I should warn you that this is a lot of physics in this video as well. If you don't like physics you can jump to the next video where I show you the result of what we're going to calculate in this video. Going back to the electron here, let's say the electron is going around counter-clockwise. The velocity of that electron at this point is tangent to the circle. That's the direction of the velocity vector. The electron has mass, m let's say. The electron is going to feel a force. It's going to be attracted to the nucleus. Opposite charges attract. This negatively charged electron is gonna feel a force towards the center of the circle. That's a centripetal force. In this case we're talking about the electric force. This is the electric force that's causing the electron to move in a circle. We can find the electric force by using Coulomb's law. Over here in the left this is Coulomb's law, the electric force is equal to K, which is a constant, times q1, which is one of the charges. Let's just say that q1 is the charge on the proton. Times the other charge, q2, which we'll say is the charge on the electron. Divided by the distance between those two charges squared. This is Coulomb's law. Let's go ahead and plug in what we know so far. K is some constant which we'll get to later. q1 I said was the charge on the proton, and the charge in the proton we'll say is e for now, so elemental charge. q2 I said was the charge in the electron, and the electron has the same magnitude of charge as the proton but it's negative. So we put in a negative e here. Divided by the distance between the two charges squared. Force is equal to mass times acceleration using Newton's second law. m is the mass of the electron. This will be the centripetal acceleration since we're talking about a centripetal force. We know that the centripetal acceleration is equal to V squared over r. We can go ahead and plug in m times v squared over r. Immediately we can cancel out one of the r's. Since we only care about the magnitude of the force we know the direction of the electric force, we don't really care about this negative sign, so we can just say we only care about the magnitude of the electric force here. We can go ahead and simplify a little bit. This would be ke squared over r on the left, and on the right this would be mv squared. Continuing with some more classical physics, next we're gonna talk about angular momentum which is a tricky concept. Angular momentum is capital L, and one equation for it is r cross p where r is a vector and p is the linear momentum. Linear momentum is equal to the mass times the velocity. We're talking about the linear momentum of the electrons, so the mass of the electron times the velocity of the electron. Let's go ahead and plug this in for angular momentum. We're gonna take the angular momentum about the center of our circle here. The angular momentum at the center, so r is a vector. Let me go ahead and draw r in. So r is a vector. It's the distance from the center to where our electron is. So we have r right there. This is the r vector. I put in r. Cross is the cross product. This would be times the linear momentum, so times p which is mv, times the sine of the angle between the two vectors. Let's think about their other vector here. The other vector is the momentum vector. We took care of the r vector. The momentum vector is in the same direction as the velocity. If this is the direction of the velocity that's also the direction of the linear momentum vector. The angle between those two vectors, the angle between those two vectors is obviously 90 degrees. Sine of 90 is 1. We can just say the angular momentum is equal to rmv times 1. Niels Bohr thought that this angular momentum should be quantized. What he did was he set this angular momentum equal to some integer, so like 1, 2 or 3, or you can keep going. But let's just say an integer n, times h which is Planck's constant divided by 2 pi. This is what Bohr came up with. He took this and he solved for the velocity. Let's go ahead and do that. On the right we're just gonna solve for v. The velocity is equal to, this would just be n times h divided by 2 pi mr. We just solve for V, and then we're gonna take that. We just solve for V and we're gonna plug that into our other equation over here on the left. Let's go ahead and do that. We would have ke squared over r. On the right we would have m times all of that, n times h over 2 pi mr. Then we just square all of that. Let's go ahead and get some more room and let's continue with our algebra here. We have ke squared over r is equal to the mass times, so we square everything in parenthesis. n squared, h squared, 4 pi squared, m squared, r squared. We can cancel a few things. We can cancel out one of these m's here. We can cancel out on of these r's. Now we would have on the left side ke squared is equal to n squared h squared over 4 pi squared. We would have one m left and one r left. The goal of all this is to solve for the radius of that circle. To solve for r we could start by multiplying both sides by 4 pi squared mr. We would get ke squared 4 pi squared mr on the left side. In the right side we would get m squared h squared. We're going to solve for r. Let's go ahead and do that. r would be equal to n squared h squared over, this would be over ke squared 4 pi squared m. Now next we're going to take all of this stuff and we are going to plug in what those numbers are. For example h is Planck's constant, we know what that is. That's 6.626 times 10 to the negative 34. We're going to be squaring that number. That's going to be over all of this. k, if you're taking physics, you know that k is equal to 9 times 10 to the ninth. It's a constant. e is elemental charge, the magnitude of charge on a proton or an electron is 1.6 times 10 to the negative 19 coulombs. We put that in there and that number needs to be squared as well. We have a 4 pi squared in there. Remember, m was the mass of the electron. You can look up the mass of the electron, it's 9.11 times 10 to the negative 31st kilograms. That's a lot of math. Rather than take out the calculator and show you, you can do that yourself. You'll see that that number comes out to be, this comes out to be, this is equal to, I'll put it down here, 5.3 times 10 to the negative 11. If you had time to do all the units you would get meters for this. Go ahead and do that calculation yourself and you would see that you get that number. That's a very important number. Let's plug that in to what we have so far in the left. The radius is equal to n squared times that number now. 5.3 times 10 to the negative 11. Let's go ahead and plug in n is equal to 1, so an integer. This represents a ground state electron in hydrogen. If n is equal to 1, this would be r1 is equal to 1 squared times this number. Obviously that's very simple math. We know that the radius, when n is equal to 1 the radius is equal to this number, 5.3 times 10 to the negative 11 meters. Let's go back up here. Let's go back up here to the picture so I can show you what we're talking about, why that's an important value. This is what we just calculated. We calculated this radius for a ground state electron in hydrogen. We calculated this distance and we called it r1. The idea of Niels Bohr by quantizing angular momentum that's going to limit your radii, the different radii that you could have. Let's go ahead and generalize this equation. We could say r, for any integer n, would be equal to n squared times this number, times r1. n squared times r1 which we just calculated to be 5.3 times 10 to the negative 11 meters. This is very important. r for any integer n is equal to n squared times r 1. This means only certain radii are allowed because Niels Bohr quantized the angular momentum. You have to have specific radii. We'll talk about the other radii in the next video. This video, after all that physics, we got this equation. We're gonna use that to go into more detail about the Bohr model radii.