Review of Bohr's model of hydrogen

As we have seen in a previous article, the emission spectra of different elements contain discrete lines. The following image shows the visible region of the emission spectra for hydrogen.

 

The quantized emission spectra indicated to Bohr that perhaps electrons could only exist within the atom at certain atomic radii and energies. Recall that quantized refers to the fact that energy can only be absorbed and emitted in a range of allowable values rather than with any possible value. The following diagram of the Bohr model shows the electron existing in a finite number of allowed orbits or shells around the nucleus.

 

From this model, Bohr derived an equation that correctly predicted the various energy levels in the hydrogen atom, which corresponded directly to the emission lines in the hydrogen spectrum. Bohr's model was also successful at predicting the energy levels in other one-electron systems, such as $\text{He}^+$He​+​​H, e, start superscript, plus, end superscript. However, it failed to explain the electronic structure in atoms that contained more than one electron.

While some physicists initially tried to adapt Bohr's model to make it useful for more complicated systems, they eventually concluded that a completely different model was needed

What have we learned since Bohr proposed his model of hydrogen?

The Bohr model worked beautifully for explaining the hydrogen atom and other single electron systems such as $\text{He}^+$He​+​​H, e, start superscript, plus, end superscript. Unfortunately, it did not do as well when applied to the spectra of more complex atoms. Furthermore, the Bohr model had no way of explaining why some lines are more intense than others or why some spectral lines split into multiple lines in the presence of a magnetic field—the Zeeman effect.

In the following decades, work by scientists such as Erwin Schrödinger showed that electrons can be thought of as behaving like waves and behaving as particles. This means that it is not possible to know both a given electron’s position in space and its velocity at the same time, a concept that is more precisely stated in Heisenberg's uncertainty principle. The uncertainty principle contradicts Bohr’s idea of electrons existing in specific orbits with a known velocity and radius. Instead, we can only calculate probabilities of finding electrons in a particular region of space around the nucleus.

The modern quantum mechanical model may sound like a huge leap from the Bohr model, but the key idea is the same: classical physics is not sufficient to explain all phenomena on an atomic level. Bohr was the first to recognize this by incorporating the idea of quantization into the electronic structure of the hydrogen atom, and he was able to thereby explain the emission spectra of hydrogen as well as other one-electron systems.

Absorption and emission

Bohr could now precisely describe the processes of absorption and emission in terms of electronic structure. According to Bohr's model, an electron would absorb energy in the form of photons to get excited to a higher energy level as long as the photon's energy was equal to the energy difference between the initial and final energy levels. After jumping to the higher energy level—also called the excited state—the excited electron would be in a less stable position, so it would quickly emit a photon to relax back to a lower, more stable energy level. 

 

The energy levels and transitions between them can be illustrated using an energy level diagram, such as the example above showing electrons relaxing back to the $n=2$n=2n, equals, 2 level of hydrogen. The energy of the emitted photon is equal to the difference in energy between the two energy levels for a particular transition. The energy difference between energy levels $n_{high}$n​high​​n, start subscript, h, i, g, h, end subscript and $n_{low}$n​low​​n, start subscript, l, o, w, end subscript can be calculated using the equation for $E(n)$E(n)E, left parenthesis, n, right parenthesis from the previous section:

$\begin{aligned} \Delta E &= E(n_{high})-E(n_{low}) \\ \\ &=\left( -\dfrac{1}{{n_{high}}^2} \cdot 13.6\,\text{eV} \right)-\left(-\dfrac{1}{{n_{low}}^2} \cdot 13.6\,\text{eV}\right) \\ \\ &= \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot 13.6\,\text{eV} \end{aligned}$​ΔE​​​​​​​=E(n​high​​)−E(n​low​​)​=(−​n​high​​​2​​​​1​​⋅13.6eV)−(−​n​low​​​2​​​​1​​⋅13.6eV)​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅13.6eV​​

Since we also know the relationship between the energy of a photon and its frequency from Planck's equation, we can solve for the frequency of the emitted photon:

$\begin{aligned} h\nu &=\Delta E = \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot 13.6\,\text{eV} ~~~~~~~~~~~~\text{Set photon energy equal to energy difference}\\ \\ \nu &= \left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h}~~~~~~~~~~~~~~~~~~~~~~\text{Solve for frequency}\end{aligned}$​hν​​ν​​​=ΔE=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅13.6eV            Set photon energy equal to energy difference​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​h​​13.6eV​​                      Solve for frequency​​

We can also find the equation for the wavelenth of the emitted electromagnetic radiation using the relationship between the speed of light $\text c$cc, frequency $\nu$ν, and wavelength $\lambda$λlambda:

$\begin{aligned}\text c &=\lambda \nu ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{Rearrange to solve for }\nu . \\ \dfrac{\text c}{\lambda}&=\nu=\left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h}~~~~~~~~~~~~~~\text{Divide both sides by c to solve for }\dfrac{1}{\lambda}.\\ \\ \dfrac{1}{\lambda} &=\left(\dfrac{1}{{n_{low}}^2}-\dfrac{1}{{n_{high}}^2}\right) \cdot \dfrac{13.6\,\text{eV}}{h\text c} \end{aligned}$​c​​λ​​c​​​​​λ​​1​​​​​=λν                                                                  Rearrange to solve for ν.​=ν=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​h​​13.6eV​​              Divide both sides by c to solve for ​λ​​1​​.​=(​n​low​​​2​​​​1​​−​n​high​​​2​​​​1​​)⋅​hc​​13.6eV​​​​

 Thus, we can see that the frequency—and wavelength—of the emitted photon depends on the energies of the initial and final shells of an electron in hydrogen.

Bohr's model of the hydrogen atom: quantization of electronic structure

Bohr’s model of the hydrogen atom started from the planetary model, 

but he added one assumption regarding the electrons. 

What if the electronic structure of the atom was quantized? Bohr suggested that the perhaps the electrons could only orbit the nucleus in specific orbits or shells with a fixed radius. Only shells with a radius given by the equation below would be allowed, and the electron could not exist in between these shells. Mathematically, we could write the allowed values of the atomic radius as $\mathrm{ <b>\; r</b>}$(n)=n2⋅r(1)r(n)=n^2\cdot r(1)r(n)=n​2​​⋅r(1)r, left parenthesis, n, right parenthesis, equals, n, start superscript, 2, end superscript, dot, r, left parenthesis, 1, right parenthesis,

where $n$nn is a positive integer, and $r(1)$r(1)r, left parenthesis, 1, right parenthesis is the Bohr radius, the smallest allowed radius for hydrogen.

He found that $r(1)$r(1)r, left parenthesis, 1, right parenthesis has the value

$\text{Bohr radius}=r\left(1\right)=0\mathrm{.}529\times 10^{-10}\text{m}$
$\text{Bohr radius}=r(1)=0.529 \times 10^{-10} \,\text{m}$By keeping the electrons in circular, quantized orbits around the positively-charged nucleus, Bohr was able to calculate the energy of an electron in the $n$nnth energy level of hydrogen: $E(n)=-\dfrac{1}{n^2} \cdot 13.6\,\text{eV}$E(n)=−​n​2​​​​1​​⋅13.6eVE, left parenthesis, n, right parenthesis, equals, minus, start fraction, 1, divided by, n, start superscript, 2, end superscript, end fraction, dot, 13, point, 6, space, e, V, where the lowest possible energy or ground state energy of a hydrogen electron—$E(1)$E(1)E, left parenthesis, 1, right parenthesis—is $-13.6\,\text{eV}$−13.6eVminus, 13, point, 6, space, e, V. Bohr radius=r(1)=0.529×10​−10​​m




Note that the energy is always going to be a negative number, and the ground state, $n=1$n=1n, equals, 1, has the most negative value. This is because the energy of an electron in orbit is relative to the energy of an electron that has been completely separated from its nucleus, $n=\infty$n=∞n, equals, infinity, which is defined to have an energy of $0\,\text{eV}$0eV0, space, e, V. Since an electron in orbit around the nucleus is more stable than an electron that is infinitely far away from its nucleus, the energy of an electron in orbit is always negative.