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atomic_transitions_and_spectroscopy [2021/02/10 19:42] – [1.viii.3 Generalizations] adminatomic_transitions_and_spectroscopy [2022/09/06 18:23] (current) – [1.viii.2 The Bohr Model of Hydrogen] admin
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 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. 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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 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
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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.