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atomic_transitions_and_spectroscopy [2021/02/09 07:42] – adminatomic_transitions_and_spectroscopy [2022/09/06 18:23] (current) – [1.viii.2 The Bohr Model of Hydrogen] admin
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 Now, accelerating charges are supposed to radiate energy, but if only a discrete set of energies are allowed for the orbits then there is no way that the electron can continuously transition from one orbit to another.  So Bohr proposed that electrons transition between orbits by jumping discontinuously and instantaneously. Now, accelerating charges are supposed to radiate energy, but if only a discrete set of energies are allowed for the orbits then there is no way that the electron can continuously transition from one orbit to another.  So Bohr proposed that electrons transition between orbits by jumping discontinuously and instantaneously.
  
-It has to be said that Bohr's model was extremely bold, by which I mean that it would have seemed completely crazy to physicists at the time.  Forget blackbody radiation, the photoelectric effect, Compton scattering and de Broglie matter waves, which only require us to do a bit of fancy footwork about particles sometimes being waves and vice versa.  Bohr is asking us to completely abandon Newtons laws and the laws of electromagnetism when it comes to electrons orbiting the nuclei of atoms.  If there is anything that indicated that a completely new theory of physics was needed it was the Bohr model of hydrogen.+It has to be said that Bohr's model was extremely bold, by which I mean that it would have seemed completely crazy to physicists at the time.  Forget blackbody radiation, the photoelectric effect, Compton scattering and de Broglie matter waves, which only require us to do a bit of fancy footwork about particles sometimes being waves and vice versa.  Bohr is asking us to completely abandon Newton's laws and the laws of electromagnetism when it comes to electrons orbiting the nuclei of atoms.  If there is anything that indicated that a completely new theory of physics was needed it was the Bohr model of hydrogen.
  
-Bohr suggested that, as long as the electron stays in one of the stationary orbits then it does not emit or absorb any electromagnetic radiation.  However, the electron may jump from a lower energy orbit $E_n$ to a higher energy orbit $E_n$, with $E_n > E_m$, by absorbing radiation with frequency $\nu$ that satisfies+Bohr suggested that, as long as the electron stays in one of the stationary orbits then it does not emit or absorb any electromagnetic radiation.  However, the electron may jump from a lower energy orbit $E_m$ to a higher energy orbit $E_n$, with $E_n > E_m$, by absorbing radiation with frequency $\nu$ that satisfies
 \[h\nu = E_n - E_m.\] \[h\nu = E_n - E_m.\]
 In other words, by absorbing a photon with energy $h\nu$. In other words, by absorbing a photon with energy $h\nu$.
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 which is also equal to $n\hbar$ by Bohr's assumption. which is also equal to $n\hbar$ by Bohr's assumption.
  
-In an in-class activity, you will be asked to take these four formulas an eliminate variables to show that the radius of the $n^{\text{th}}$ orbit is +In an in-class activity, you will be asked to take these four formulas and eliminate variables to show that the radius of the $n^{\text{th}}$ orbit is 
-\[r_n = \frac{4\pi\epsilon_0\hbar^2}{me^2} = a_0 n^2,\]+\[r_n = \frac{4\pi\epsilon_0\hbar^2}{me^2}n^2 = a_0 n^2,\]
 where where
 \[a_0 = \frac{4\pi\epsilon_0\hbar^2}{me^2},\] \[a_0 = \frac{4\pi\epsilon_0\hbar^2}{me^2},\]
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 Substituting the expressions for $r_n$ and $v_n$ into the energy $E_n$ and simplifying (which I will not subject you to doing in an in-class activity) gives Substituting the expressions for $r_n$ and $v_n$ into the energy $E_n$ and simplifying (which I will not subject you to doing in an in-class activity) gives
-\[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0} right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}}{n^2}},\]+\[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0} \right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}}{n^2}},\]
 which is known as the //**Bohr Energy**//. which is known as the //**Bohr Energy**//.
  
 The constant $\mathcal{R}$ is the magnitude of the energy of the $n=1$ orbit and is known as the //**Rydberg Energy**//.  Its value is The constant $\mathcal{R}$ is the magnitude of the energy of the $n=1$ orbit and is known as the //**Rydberg Energy**//.  Its value is
 \[\boxed{\mathcal{R} = \frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0}\right )^2 \approx 13.6\,\text{eV}.}\] \[\boxed{\mathcal{R} = \frac{m_e}{2\hbar^2} \left ( \frac{e^2}{4\pi\epsilon_0}\right )^2 \approx 13.6\,\text{eV}.}\]
 +
 +Note: the fact that there is a lowest possible energy for the Bohr atom and that this corresponds to a finite radius and finite speed means that the electron does not spiral into the nucleus radiating ever higher frequency radiation as it does so.  It has to stop at the lowest energy state, so matter is stable and the universe is saved from disintegration.  Hooray!
  
 The set of allowed energies for a hydrogen atom is depicted below. The set of allowed energies for a hydrogen atom is depicted below.
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 ====== 1.viii.3 Generalizations ====== ====== 1.viii.3 Generalizations ======
  
-Right, now that we have the hydrogen atom down, let's do helium and the rest of the periodic table!  Sadly, things are not so simple.  As soon as we have two or more electrons, we have to take into account the electrostatic repulsion between the electrons.  The Bohr model does not work because we can no longer reasonably expect the orbits to be circular.  Even in full-blown quantum mechanics, we cannot even solve the helium atom without approximations.  So helium, and the rest of the periodic table, will have to wait until PHYS 452.+Right, now that we have the hydrogen atom down, let's do helium and the rest of the periodic table!  Sadly, things are not so simple.  As soon as we have two or more electrons, we have to take into account the electrostatic repulsion between the electrons.  The Bohr model does not work because we can no longer reasonably expect the orbits to be circular.  Even in full-blown quantum mechanics, we cannot solve the helium atom without approximations.  So helium, and the rest of the periodic table, will have to wait until PHYS 452.
  
 Your chemistry professors may wonder why us physicists spend an entire course attempting to solve the hydrogen atom, only obtaining the full quantum mechanical model at the very end, when there are so many other interesting atoms and molecules out there.  Us physicists like to understand the simplest system in full detail before moving on, and the hydrogen atom is our "quantum spherical cow". Your chemistry professors may wonder why us physicists spend an entire course attempting to solve the hydrogen atom, only obtaining the full quantum mechanical model at the very end, when there are so many other interesting atoms and molecules out there.  Us physicists like to understand the simplest system in full detail before moving on, and the hydrogen atom is our "quantum spherical cow".
  
-So, what can we do that is more general than hydrogen at this point.  Well, we can certainly deal with ions that have heavier nuclei than hydrogen, but have lost all of their electrons except one.  These are known as //hydrogen-like// ions.  The only effect this has on our calculations is that the charge on the proton $e$ is replaced by the charge on the heavier nucleus $Ze$ (where $Z$ is the //atomic number//, i.e. the number of protons in the nucleus).  This means that one of the factors of $e$ in all our formulas will be replaced by $Ze$.  The other factor of $e$ does not change because that comes from the charge of the electron.+So, what can we do that is more general than hydrogen at this point?  Well, we can certainly deal with ions that have heavier nuclei than hydrogen, but have lost all of their electrons except one.  These are known as //hydrogen-like// ions.  The only effect this has on our calculations is that the charge on the proton $e$ is replaced by the charge on the heavier nucleus $Ze$ (where $Z$ is the //atomic number//, i.e. the number of protons in the nucleus).  This means that one of the factors of $e$ in all our formulas will be replaced by $Ze$.  The other factor of $e$ does not change because that comes from the charge of the electron.
  
 So, for a hydrogen-like ion, the Bohr energy becomes So, for a hydrogen-like ion, the Bohr energy becomes
-\[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{Ze^2}{4\pi\epsilon_0} right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}Z^2}{n^2}},\]+\[\boxed{E_n = -\frac{m_e}{2\hbar^2} \left ( \frac{Ze^2}{4\pi\epsilon_0} \right )^2 \frac{1}{n^2} = -\frac{\mathcal{R}Z^2}{n^2}},\]
  
 The other thing we could do is remove the approximation we made that the nucleus remains stationary.  It is, in fact, a really good approximation because the mass of the proton is $m_pc^2 = 938\,\text{MeV}$, so the ratio of the electron mass to the proton mass is $m_e/m_p = 0.511/938 \approx 5\times 10^{-4}$.  However, it is possible to make atom-like systems by replacing the proton by a positively charged particle with much smaller mass that is closer to the mass of an electron.  If you go on to study physics in graduate school you will learn that every particle has an anti-particle that has the same mass but opposite charge.  The anti-particle of the electron is called the positron and physicists have managed to make a bound system of an electron and positron, which is called //positronium//. The other thing we could do is remove the approximation we made that the nucleus remains stationary.  It is, in fact, a really good approximation because the mass of the proton is $m_pc^2 = 938\,\text{MeV}$, so the ratio of the electron mass to the proton mass is $m_e/m_p = 0.511/938 \approx 5\times 10^{-4}$.  However, it is possible to make atom-like systems by replacing the proton by a positively charged particle with much smaller mass that is closer to the mass of an electron.  If you go on to study physics in graduate school you will learn that every particle has an anti-particle that has the same mass but opposite charge.  The anti-particle of the electron is called the positron and physicists have managed to make a bound system of an electron and positron, which is called //positronium//.
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 which you should be familiar with from your classical mechanics courses. which you should be familiar with from your classical mechanics courses.
  
-In full generality, if we have a hydrogen-like ion where the "nucleus" is made of $Z$ particles each with charge $e$ and the total mass of the nucleus is $M$, then the radius of the $n^{\text{th}}$ orbit becomes+In full generality, if we have a hydrogen-like ion where the "nucleus" is has $Z$ positively charged particles, each with charge $e$, and the total mass of the nucleus is $M$, then the radius of the $n^{\text{th}}$ orbit becomes
 \[r_n = \frac{4\pi\epsilon_0 \hbar^2}{\mu Ze^2}n^2 = \left ( 1 - \frac{m_e}{M}\right )\frac{a_0}{Z}n^2,\] \[r_n = \frac{4\pi\epsilon_0 \hbar^2}{\mu Ze^2}n^2 = \left ( 1 - \frac{m_e}{M}\right )\frac{a_0}{Z}n^2,\]
 and the energy of the $n^{\text{th}}$ orbit becomes and the energy of the $n^{\text{th}}$ orbit becomes
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   * For $m=1$, the atom drops into its ground state and emits ultraviolet radiation.  This is called the //Lyman series// and has $h\nu_L = \mathcal{R}\left ( 1 - \frac{1}{n^2} \right )$.   * For $m=1$, the atom drops into its ground state and emits ultraviolet radiation.  This is called the //Lyman series// and has $h\nu_L = \mathcal{R}\left ( 1 - \frac{1}{n^2} \right )$.
   * For $m=2$, the atom drops into its first excited state and emits visible light.  This is called the //Balmer Series// and has $h\nu_B = \mathcal{R} \left ( \frac{1}{4} - \frac{1}{n^2}\right )$.   * For $m=2$, the atom drops into its first excited state and emits visible light.  This is called the //Balmer Series// and has $h\nu_B = \mathcal{R} \left ( \frac{1}{4} - \frac{1}{n^2}\right )$.
-  * For $m=3$, the atom drops into its second excited state and emits infra-red light.  This is called the //Paschen Series// and has $h\nu_B = \mathcal{R} \left ( \frac{1}{9} - \frac{1}{n^2}\right )$.+  * For $m=3$, the atom drops into its second excited state and emits infra-red light.  This is called the //Paschen Series// and has $h\nu_P = \mathcal{R} \left ( \frac{1}{9} - \frac{1}{n^2}\right )$.
  
 Are you getting bored of this yet?  Some of the other series with higher values of $m$ also have fancy names.  I promise I will never make you remember these names for an exam.  In an ideal world, we would just call them the "$m=1$ series", the "$m=2$ series", etc.  I blame chemists for the fact that the fancy names have stuck around.  They are the people who most often do experimental spectroscopy and they need to have complicated jargon to make it look like their subject is difficult.  I often wonder if they actually understand the simple physics underlying the experiments they are doing.  I am not posting this in public am I?  Please don't tell your chemistry professors that I said any of that. Are you getting bored of this yet?  Some of the other series with higher values of $m$ also have fancy names.  I promise I will never make you remember these names for an exam.  In an ideal world, we would just call them the "$m=1$ series", the "$m=2$ series", etc.  I blame chemists for the fact that the fancy names have stuck around.  They are the people who most often do experimental spectroscopy and they need to have complicated jargon to make it look like their subject is difficult.  I often wonder if they actually understand the simple physics underlying the experiments they are doing.  I am not posting this in public am I?  Please don't tell your chemistry professors that I said any of that.