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## The Bohr model of the atom

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.

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.

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.

### 3 Responses

1. on 21/11/2013 at 08:28 | Reply Hipster goalies – Niels & Harold Bohr | No Standing

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2. on 02/12/2013 at 18:00 | Reply Niels Bohr Danish Physicist Who Was Credited With Establishment of Cern | Ace History News

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3. […] in 1913 that electrons could only occupy certain orbits. I go into the details of his argument in this blog, but to summarise it briefly here, he suggested that something called the orbital angular momentum […]