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Posts Tagged ‘Ernest Rutherford’

In my blog on Niels Bohr, number 2 in The Guardian’s list of the 10 best physicists, I mentioned the Bohr model of the atom. In this blog I will go into more detail about this model, and how it agreed with the experimental results (the Rydberg formula) for the hydrogen atom.

Quantised orbits

In 1911, Rutherford had proposed that atoms have positively charged nuclei, with the negatively charged electrons orbiting the nuclei. One of the problems with this idea was that an orbiting electron would be accelerating, by virtue of moving in a circle. The acceleration is directed towards the centre of the circle. It was well known that when an electron is accelerated it radiates electromagnetic waves. Calculations showed that the orbiting electrons on Rutherford’s model should radiate away their energy in a few microseconds (millionths of a second), and spiral towards the nucleus. They clearly were not doing this, but why?

Bohr suggested in a paper in 1913 that electrons would somehow not radiate away their energy if they were orbiting in certain “allowed orbits”. If they were in these special orbits, the normal laws of EM radiation would not apply. He suggested that these allowed orbits were when the orbital angular momentum L could be written as


L = \frac{ n h }{ 2 \pi } = n \hbar \text{  (Equ. 1) }


(\hbar = \frac{ h }{ 2 \pi } \text{ where } h is Planck’s constant, and is given its own symbol in Physics at it crops up so often). What is orbital angular momentum? Well, it is the rotational equivalent of linear momentum. Linear momentum is defined as \vec{p} = m\vec{v} \text{ where } m \text{ is the mass and } \vec{v} \text{ is the velocity}. Notice, momentum is a vector quantity, this is important in doing calculations involving collisions, such as the ones I did in this blog.



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By analogy, orbital angular momenutm is defined as


\vec{L} = \vec{r} \times m \vec{v}


where \vec{r} is the radius vector of the orbit, which is defined as pointing from the centre of the orbit along the radius. For a circular orbit, where the radius vector is at right angles to the velocity vector, we can just write L=mvr where L is the magnitude (size) of the vector \vec{L}.

The force keeping the electron in orbit

From Classical Physics, Bohr argued that the force which was keeping the electron in orbit about the positively charged nucleus was the well known Coulomb force, given by


F = - \frac{ Z k_{e} e^{2} }{ r^{2} }


where k_{e} is the Coulomb constant (which determines the force between two 1 Coulomb charges separated by 1 metre), Z is the atomic number of the atom, e is the charge on an electron and r is the radius of the orbit. The minus sign is telling us that the force is directed towards the centre, whereas our definition of the radius vector is that it is away from the centre, so they are in opposite directions.

We can equate this to the formula for the centripetal force on any object moving in a circular orbit, so we can write


\frac{ m_{e} v^{2} }{ r } = \frac{ Z k_{e} e^{2} }{ r^{2} } \text{ (Equ. 2) }


where m_{e} is the mass of the electron and v is the speed of its orbit.

The Bohr radius

Re-arranging Equation 2 we can write


v = \sqrt{ \frac{ Z k_{e} e^{2} m_{e} r }{ m_{e}^2 r^{2} } }


which then allows us to write the angular momentum as


m_{e} v r = \sqrt{ Z k_{e} e^{2} m_{e} r } \text{  which (from Equ. 1) } =  n \hbar


This allows us to write an expression for the “radius” of an electron’s orbit as


r_{n} = \frac{ n^{2} \hbar^{2} }{ Z k_{e} e^{2} m_{e} }


where n is the energy level of the electron. The so-called “Bohr radius” is the radius of an electron in the n=1 energy level for hydrogen (Z=1) and can be written


\boxed{ r_{1} = \frac{ \hbar^{2} }{ k_{e} e^{2} m_{e} } \approx 5.29 \times 10^{-11} \text{ metres } }


This is, indeed, about the size of a hydrogen atom.

The total energy of the electron

The total energy of the electron in its orbit is given by the sum of its kinetic energy and its potential energy. The kinetic energy is just given by 1/2 \; (mv^{2}). What about the potential energy? The potential energy can be found by using the relationship between work and force; back in this blog I said that work was defined as the force multiplied by the distance moved. Energy is the capacity to do work, and is measured in the same units, Joules. So we can derive the potential energy of an electron in orbit due to the Coulomb force as


P.E. = \int_{r}^{\infty} { F} dr = - \int_{r}^{\infty} \frac{ Z k_{e} e^{2} }{ r^{2} } dr


where dr is an incremental change in the radius. If we do this integration we get


P.E. = - \frac{ Z k_{e} e^{2} }{ r }


where the negative sign is telling us that we have to do work on the electron to increase its radius, or to put it another way that the force acts towards the centre but the radius vector acts away from the centre of the electron’s orbit. This means that the total energy E is given by


E = \frac{ 1 }{ 2 } m_{e} v^{2} - \frac{ Z k_{e} e^{2} }{ r }


But, from Equ. 2 we can write the kinetic energy as


\frac{ 1 }{ 2 } m_{e} v^{2} = \frac{ Z k_{e} e^{2} }{ 2r }


So then the total energy E can be written


E = \frac{ Z k_{e} e^{2} }{ 2r } - \frac{ Z k_{e} e^{2} }{ r } = - \frac{ Z k_{e} e^{2} }{ 2 r }
So, in the Bohr model, the energy of the n^{th} energy level is given by


\boxed{ E_{n} = - \frac{ Z k_{e} e^{2} }{ 2 r_{n} } \text{ or } -\frac{ Z^{2} (k_{e} e^{2})^{2} m_{e} }{2 \hbar^{2} n^{2} }  }


In the case of hydrogen, where Z=1 we can write


\boxed{ E_{n} =  -\frac{ (k_{e} e^{2})^{2} m_{e} }{ 2 \hbar^{2} n^{2} } \approx -\frac{ 13.6 }{ n^{2} } \text{ eV} }


This was in perfect agreement with the Rydberg formula for the energy levels of hydrogen, which had been experimentally derived by the Swedish physicist Johannes Rydberg in 1888. As I will show in a future blog, Bohr’s model was a “semi-empirical” model, in that it was a step along the way to the correct model. It was produced by using a mixture of classical physics and quantum mechanics, and Bohr did not understand why his condition that only orbits whose angular momentum were equal to n \hbar was true. The explanation was produced with the full theory of Quantum Mechanics in 1926 as a solution to Schrödinger’s wave equation for hydrogen.

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At number 2 in The Guardian’s list of the ten best physicists is Niels Bohr.

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Bohr is best known for his work on the allowed orbits an electron in an atom can have. This is the so-called “Bohr model”, and in 1922 he won the Nobel Prize in Physics for this suggestion that electrons can only exist in certain allowed orbits, which naturally explained the line spectra of the elements.

Bohr’s brief biography

Niels Bohr was born in Copenhagen in 1885, the second of three children. His father, Christian Bohr, was a professor of physiology at the University of Copenhagen. His mother, Ellen Adler, came from a wealthy Danish Jewish family. At 18, Bohr enrolled at Copenhagen University where he studied physics. In 1905, whilst still an undergraduate, Bohr won the first prize in a competition sponsored by the Royal Danish Academy of Arts and Sciences. In May 1911 he obtained his PhD for a thesis on the electron theory of metals.

In the same year, Bohr took up a post-doctoral research position at The University of Manchester, working in Ernest Rutherford’s research team. He arrived in Manchester just as Rutherford was proposing the theory that atoms contained small, positively charged nuclei; where nearly all the atom’s mass was concentrated. However, by 1912 Bohr had returned to his native Denmark where he got a job teaching medical students. In 1913 he published his first paper suggesting what is now known as the “Bohr model” of the atom.

He returned to Manchester in 1914, and spent two more years working with Rutherford, as a Reader in the Physics Department. Then, in 1916 a Professorship in Theoretical Physics was created for him at the University of Copenhagen. In 1918 Bohr started trying to establish an institute of theoretical physics in Copenhagen, the institute opened in 1921 and became known as the Niels Bohr Institute. Apart from fleeing from Nazi occupied Denmark in 1943, Bohr spent the rest of his career as the Director of this institute he had established, and died in 1962 at the age of 77.

Bohr’s contributions to Physics

When Rutherford proposed his model of the atom with a positively charged nucleus and the electrons in orbit about it, a problem arose. Classical physics predicted that an electron in orbit, because it is constantly accelerating through changing its direction, should be constantly radiating. As a consequence, it should lose its energy and spiral in towards the nucleus. Calculations showed that this should happen in millionths of a second, meaning all atoms would be unstable.

In 1913, Bohr suggested that electrons could only exist in certain allowed orbits. He suggested that these orbits were “quantised”, and that the angular momentum of electron orbits had to be equal to L=n\hbar where n is an integer (1,2,3 etc.) and \hbar \text{ is } h/2\pi where h is Planck’s constant.

Bohr then suggested that electrons could be excited from one orbit to another, either by gaining some (or all) of the energy of an incoming electron, or by absorbing all of the energy of an incoming photon. When electrons were excited to a higher energy level, they would quickly jump back down to the lowest available orbit, and in so doing would emit light (photons) of particular wavelengths.

This “Bohr model” was able to naturally explain the observed spectra of hydrogen (which only has one electron and one proton), and of singly-ionised helium, which again only has one electron. Its success led to Bohr being awarded the Nobel Prize in 1922, the citation reading “for his services in the investigation of the structure of atoms and of the radiation emanating from them”.

Bohr played a central role in the probabilistic interpretation of Quantum Mechanics. The so-called “Copenhagen interpretation”, of which he was the main champion, was that physics was not able, under the laws of quantum mechanics, to give us any more than the probabilities of the outcomes of experiments. This was in stark contrast to e.g. Einstein, who believed that that nature was inherently deterministic not probabilistic.

Bohr was instrumental in the establishment of CERN, the European Centre for Particle Physics Research. He was very much one of the elder-statesmen of the Physics community, and a period at his Institute in Copenhagen became almost essential in the career of any theoretical physicist.

Do you think Niels Bohr deserves to be in this list of the ten best physicists?

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You can read more about Niels Bohr and the other physicists in this “10 best” list in our book 10 Physicists Who Transformed Our Understanding of the UniverseClick here for more details and to read some reviews.

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Ten Physicists Who Transformed Our Understanding of Reality is available now. Follow this link to order

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I was in Cambridge in early February with my son, and on our 3rd day there we went out to the Cavendish Laboratory and the Institute of Astronomy. In the road that runs from the main road to the Cavendish is JJ Thomson Avenue. As this picture shows, the avenue is named after the famous English physicist J.J. Thomson.


The sign on JJ Thomson Avenue, the road which leads to the Cavendish laboratories in Cambridge

The sign on JJ Thomson Avenue, the road which leads from the main road to the Cavendish laboratories in Cambridge


In 1897 J.J. Thomson discovered the first sub-atomic particle, the electron. He was doing experiments with cathode ray tubes, which had been discovered in the mid 1880s, and decided to see how they might be affected by magnetic and electric fields. He found that the cathode rays were deflected by the magnetic and electric fields, showing that they must be made up of charged particles.


Two frames of reference S and S' moving relative to each other have a flash of light originate at their respective origins at time t=0

A cathode ray tube. Electrons travel from the negative end (the cathode) towards the positive terminal (the anode). The green glow is produced by the charged electrons interacting with phosphorous in the glass, which then fluoresces. The shadow of the Maltese cross appears on the glass.


More than this, be was able to measure the shape of curve produced by the fields, and using the known strength of the fields he used, was able to calculate the mass and charge of the particles. By this time, chemists had fairly accurately determined the masses of atoms, and had shown that e.g. Carbon had a mass some 12 times the mass of Hydrogen.

Thomson found that the mass of the particles he was deflecting in his magnetic and electric fields werethousands of times less massive than the mass of the lightest known element, Hydrogen. This of course indicated that what he had discovered was sub-atomic, a constituent of atoms.

Thomson thus showed that cathode rays are a stream of electrons. The very high voltage (thousands of volts) between the positive (anode) and negative (cathode) terminals causes electrons in the cathode to be accelerated to a high velocity as they are attracted towards the anode. Cathode ray tubes were the basis for most TVs and computer monitors until this last 10 years, when more efficient Liquid Crystal Displays have largely replaced them.

As I will describe in a future blog in more details, some 12-13 years after Thomson’s work, one of his ex-students Ernest Rutherford showed that most of the mass of an atom resides in the centre, in its nucleus. The electrons orbit the nucleus, but it is the electrons which are important in e.g. giving elements their chemical properties and forming bonds with other atoms.

One anecdote about Thomson is that he used to get annoyed when people would write to him and spell his name with the more traditional “p” in it, Thompson. In replying to any such correspondence, he would insert random letter “p”s into the person’s name.

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In a previous blog I mentioned that I would derive some equations relating to two body collisions, which would, among other things, allow one to see how quickly (and hence how far) one can hit a golf ball with a golf driver. I do this in this blog. I am sorry for it being so long, and also so mathematical, but there is really no other way to derive these important relationships than to go through the mathematics.

Conservation of momentum and elastic collisions

In all collisions, momentum, defined as \vec{p} = m\vec{v} where m is the mass and \vec{v} is the velocity, is conserved. Momentum is a vector, so has both magnitude (size) and direction.

In an elastic collision, kinetic energy is also conserved. Kinetic energy is the energy of motion, and is defined as KE=\frac{1}{2}mv^{2}. We will consider two masses colliding, the mass of object 1 is m_{1}, the mass of object 2 is m_{2}. The velocity before the collision of object 1 is u_{1}, and of object 2 is u_{2}, the velocity after the collision of object 1 is v_{1} and that of object 2 is v_{2}.

The conservation of momentum allows us to write m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} and the conservation of kinetic energy allows us to write \frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m_{2}u_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}

Collisions where object 2 is stationary before the collision

In this blog, I am going to consider the case of object 2 having no velocity before the collision. This is, in fact, very common. For example, it is the sort of collision we have when Tiger Woods’ golf driver hits his golf ball as he drives it up the fairway (or when his ex-wife allegedly hit his car with a golf club when she learned of his affairs!).

Tiger Woods driving his golf ball up the fairway is an example of a two body collision

The expression for the velocity of object 1 v_{1} after the collision

If object 2 is stationary before the collision then u_{2}=0 and so we can write, for the conservation of momentum, m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2} \qquad(1) and, for conservation of kinetic energy, \frac{1}{2}m_{1}u_{1}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} \qquad(2).

Re-arranging equation (1) we get m_{2}v_{2} = m_{1}u_{1} - m_{1}v_{1} = m_{1}(u_{1}-v_{1}) and so v_{2} = \frac{m_{1}(u_{1}-v_{1})}{m_{2}} \qquad(3).

Multiplying each term in equation (2) by 2, and re-arranging, we get m_{2}v_{2}^{2} = m_{1}u_{1}^{2} - m_{1}v_{1}^{2} = m_{1}(u_{1}^{2} - v_{1}^{2}) and so v_{2}^{2} = \frac{m_{1}(u_{1}^{2} - v_{1}^{2})}{m_{2}} = \frac{m_{1}(u_{1}+v_{1})(u_{1}-v_{1})}{m_{2}} \qquad(4)

If we now square equation (3) we get v_{2}^{2} = \frac{m_{1}^{2}(u_{1}-v_{1})(u_{1}-v_{1})}{m_{2}^{2}} \qquad (5)

We can set this equal to equation (4), which allows us to write \frac{m_{1}^{2}(u_{1}-v_{1})(u_{1}-v_{1})}{m_{2}^{2}} = \frac{m_{1}(u_{1}+v_{1})(u_{1}-v_{1})}{m_{2}} . Cancelling common terms on both sides, this becomes \frac{m_{1}(u_{1}-v_{1})}{m_{2}} = u_{1}+v_{1} which can then be re-arranged to give m_{1}(u_{1}-v_{1}) = m_{2}(u_{1}+v_{1}). Re-arranging this to get the terms involving v_{1} all on one side, m_{1}v_{1}+m_{2}v_{1} = m_{1}u_{1}-m_{2}u_{1} and so we can write v_{1}(m_{1}+m_{2}) = u_{1}(m_{1}-m_{2}).

So, finally, the expression for the velocity of object 1 after the collision is \boxed { v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} }

The expression for the velocity of object 2 after the collision, v_{2}

To find v_{2} we need to substitute this value for v_{1} from equation (3), so we can write m_{2} v_{2} = m_{1} u_{1} - m_{1}v_{1} = m_{1} u_{1} - m_{1} \left ( \frac{u_{1} (m_{1} -m_{2})} {(m_{1} + m_{2}) } \right )
which can then be written m_{2} v_{2} = \frac{ m_{1} u_{1} ( m_{1} + m_{2} ) - m_{1} u_{1} ( m_{1} - m_{2} ) } { m_{1} + m_{2} }

Which becomes m_{1} v_{2} ( m_{1} + m_{2} ) = m_{1}^{2} u_{1} + m_{1} m_{2} u_{1} - m_{1}^{2} u_{1} + m_{1} m_{2} u_{1}

which becomes v_{2} ( m_{1} + m_{2} ) = 2 m_{1} u_{1}

So, finally, the expression for the velocity of object 2 after the collision is \boxed{ v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } }

The case when m_{1}<;<;m_{2}

When a moving object strikes a stationary object which is much more massive, what will happen? Intuition tells us that the stationary object will not move, and that the incoming moving object will rebound. For example, this is what happens when a ball hits a wall.

The equations for v_{1} and v_{2} which we derived above can show us that this is indeed true. If we make $m_{1} <;<; m_{2}$ in the equation for v_{1} we get v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} \approx \frac{u_{1}(-m_{2})}{m_{2}} \approx -u_{1}. Remembering that, in fact, v_{1} and u_{1} are vectors, so in fact we should write \boxed{\vec{v_{1}} \approx -\vec{u_{1}}}, so the velocity after the collision with a stationary, more massive object, is that the object returns in the opposite direction to its initial direction, but with the same speed.

When an object strikes a stationary object, we can also work out the conditions necessary for the incoming object to rebound. In other words, the conditions for v_{1} to be negative. One can see that all that is required for v_{1} to be negative is that m_{2} >; m_{1}. So, whenever a less massive object strikes a stationary more massive object, the incoming less massive object will rebound.

In the famous Geiger-Marsden experiment, performed at the University of Manchester in 1909 by Hans Geiger and Ernest Marsden, under the direction of Ernest Rutherford, they found that some alpha-particles fired at a thin sheet of gold foil were bouncing back towards the source. It is because of the equation above for v_{1} that Rutherford knew the alpha-particles must be striking something more massive, which he correctly concluded was a densely packed nucleus in the gold atoms.

The case when m_{1} = m_{2}

What about when the mass of the two objects is the same? In this case, we can get an idea from Newton’s cradle, as shown in my blog of a few weeks ago, and illustrated in this video when one ball hits the other 4 stationary balls.

What happens is that the incoming ball transfers all its momentum to the stationary ball, so that the first ball stops and the second ball moves in the same direction as the original ball, and with the same speed. If we let m_{1}=m_{2}=m then, mathematically, we can see that, for v_{1}

v_{1} we get v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} = \frac{u_{1}(m-m)}{(m+m)} = 0

so object 1 does, indeed, come to rest after the collision. But, for object 2 we can write

v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } = \frac{ 2m u_{1} }{2m} = u_{1}

so object 2 moves off after the collision with the same velocity that object 1 had before the collision. This can all be seen in the video above.

The case when m_{1} >;>; m_{2}

What about when the mass of the incoming object is much greater than the mass of the stationary object? In this case, the equation for v_{1} becomes

v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} \approx \frac{u_{1}m_{1}}{m_{1}} \approx u_{1}

so the 1st object carries on after the collision with the same velocity it had before the collision, as if the less massive object were not there.

For the initially stationary less massive object, we can calculate

v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } \approx \frac{2m_{1}u_{1}}{m_{1}} \approx 2u_{1}

which may come as a surprise to you. The initially stationary object will move off after the collision with twice the velocity of the incoming more massive object.

Applying these equations to golf

When Tiger Woods is trying to drive his golf ball up the fairway, he is of course not using a club head which is infinite in mass. The typical mass of a golf ball is probably about 40 grams, with the maximum allowed being 45.9 grams. The mass of the head of a golf club seems to be less regulated, I cannot find any regulations governing the mass of a golf club head, although I did find some regulations about the volume. But, I did find that the typical mass of a golf driver head is about 200g.

So, for simplicity, less us assume the mass of the golf driver head is 4 times the mass of the golf ball, which is approximately true. That is, m_{1}=4m_{2}. Then, we can show that the maximum value for v_{2}, the velocity of the golf ball after it has been hit, is

v_{2} = \frac{2m_{1}u_{1}}{m_{1}+m_{2}} = \frac{8m_{2}u_{1}}{5m_{2}} = \frac{8u_{1}}{5}

so the maximum velocity the golf ball can have (and hence how far it can go) is 1.6 u_{1}, the velocity of the golf club head. The faster one can swing the club, the further the ball will go, but with these figures where m_{1} = 4m_{2}, it can never acquire a velocity which is more than 1.6 times the velocity of the golf club head.

In a (much shorter) future blog, I will explain why it would not necessarily help to increase the golf club head mass, even though in theory one could get close to having v_{2} to be 2u_{1} rather than the 1.6 u_{1} we get here.

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