Bohr Atom
Neils Bohr tried to understand hydrogen (the simplest atom). He postulated the existence of certain stationary states. If the electron is orbiting in one of these states, the atom is stable. Each of these states has a particular energy. If the energy of the nth state is En and the energy of the mth state is Em, and En > Em.
Under the right conditions, transitions between states can take place. If the electron is in the higher energy state, it can drop down to the lower energy state, as long as a photon is emitted with an energy equal to the energy difference between the two states. If the frequency of the photon is ν, this means that:
\[h\nu =E_n -E_m\]
If the electron is in the lower energy state, it can make a transition to the upper state if the atom absorbs a photon with exactly the right energy. This explanation incorporated Einstein’s idea of photons.
Bohr pointed out that one could calculate the energies of the allowed states by assuming that the angular momentum J of the orbiting electrons is quantized in integer multiples of h/2π. The combination h/2π or simply \(h\bar\). Applying this to a hydrogen atom, with an electron, with charge −e, orbiting a distance r from a nucleus with charge +e. Assuming that the nucleus is much more massive than the electron, so the small motion of the nucleus, can be ignored since both the nucleus and electron orbit their common center of mass.
We first look at the kinetic energy.
\[E_k=\frac 12 mv^2\]
The potential energy, relative to the potential energy being zero when the electron is infinitely far from the nucleus, is given by:
\[E_p=\frac {-e^2}{r}\]
(Writing the potential energy in this form, rather than with a factor of \(1/4\pi \epsilon_0\), indicates the usage of cgs units. This means that charges are expressed in electrostatic units, esu, with the charge on the electron being 4.8 10^−10 esu.)
The total energy:
\[E=\frac12 mv^2 -\frac {e^2}{r}\]
v and r are relating, noting that the electrical force between the electron and the nucleus (Coulomb Force), \(e^2/r^2\), must provide the acceleration to keep the electron in the circular orbit (Centrifugal Force), \(mv^2/r\):
\[\frac {mv^2}{r}=\frac {e^2}{r^2}\]
\[mv^2=\frac {e^2}{r}\]
So:
\[E=\frac 12 \frac {e^2}{r} -\frac {e^2}{r}\]
\[E=-\frac 12 \frac {e^2}{r}\]
The minus sign indicates that the total energy is negative. The potential energy is defined zero when the electron and proton are infinitely far apart. The electron and proton being far apart with no motion is the minimal condition for the electron being free of the proton. So, if the electron is barely free of the proton, the total energy would be zero. So, if the total energy is negative, as in the equation, then it must be added energy [(1/2)e2/r] to bring it up to zero, which would free the electron. So the negative energy means that the electron is not free. In this case the system is bound.
To find the allowed values of r. The angular momentum is J = mvr. The quantization condition becomes:
\[mvr=n \frac h{2\pi}\]
\[v=n \frac h{2\pi mr}\]
Squaring the equation:
\[v^2=n^2 \frac {h^2}{4\pi ^2 m^2r^2}\]
Mltuipling by m:
\[mv^2=n^2 \frac {h^2}{4\pi ^2 mr^2}\]
Putting \(mv^2=\frac {e^2}{r}\):
\[\frac {e^2}{r}=n^2 \frac {h^2}{4\pi ^2 mr^2}\]
\[r=n^2 \frac {h^2}{4\pi ^2 me^2}\]
Giving the radius of the nth orbit:
\[r_n=n^2 \frac {h^2}{4\pi ^2 me^2}\]
Subsituting \(r_n\)in \(E=-\frac 12 \frac {e^2}{r} \)gives the energy of the nth state:
\[E=-\frac {2\pi^2me^4}{n^2 h^2}\]
Calculation of \(-\frac {2\pi^2me^4}{h^2}\) gives it the value -13.6 eV:
\[E=\frac {-13.6}{n^2}\ \ eV\]
If n = ∞ in equation, the total energy becomes E = 0. However, n = ∞ corresponds to a free electron (energy level). Therefore, to move the electron far from the nucleus (called the binding energy), it must be added 13.6 eV.
The energy goes to do work against the electrical attraction between the electron and the nucleus as the electron being pulled away.
This has the 1/n2 dependence that is expected from the Balmer formula.
Derivation the Balmer formula:
The equation \(E=-\frac {2\pi^2me^4}{n^2 h^2}\) can be writen:
\[E_n=-\frac {hcR}{n^2}\]
\[R=\frac {2\pi^2 me^4}{ch^3}\]
The energy of an emitted or absorbed photon must equal the energy difference between the two states:
\[E_n-E_m=h\nu =\frac {hc}\lambda\]
\[-\frac {hcR}{n^2}+\frac {hcR}{m^2}=\frac {hc}{\lambda}\]
\[R(\frac {1}{m^2}-\frac {1}{n^2})=\frac {1}{\lambda}\]
It looks very similar to the Balmer formula, except that the Balmer formula has a 2 instead of the m. This means that the Balmer series all have the second energy level as their lower level.
This equation can be used to divide the hydrogen spectrum into different series. A given series is characterized by having the same lower energy state. For example, the Balmer series consists of absorptions accompanying transitions from level 2 to any higher levels, and emissions accompanying transitions from higher levels down to level 2. The first Balmer transition (involving levels 2 and 3) has the smallest energy difference of the series. (Clearly the energy difference between levels 2 and 3 is less than the energy difference between levels 2 and 4, or between levels 2 and 5, and so on.) The Balmer series is important because the first few transitions fall in the visible part of the spectrum. The series with the lower energy level being level 1 is called the Lyman series. Even the lowest transition in the Lyman series is in the ultraviolet.
Transition Labeling System:
There is a labeling system for various transitions.
- Chemical symbol for denoting the element.
- m for the lowest level that characterizes the series (1 for Lyman, 2 for Balmer, etc.).
- A Greek letter denoting the number of levels jumped. For example, (α) transition for n = m + 1, (β) transitionfor n = m + 2.
The first Balmer line is then designated H2α. (Note that for the Balmer series of hydrogen only, we sometimes drop the 2 and just say Hα, Hβ, etc.)
One modification that should be made is to account for the motion of the nucleus (since it is not infinitely massive). The mass of the electron, m, in equations should be replaced by the reduced mass of the electron and proton. The reduced mass, m_r, is defined such that the motion of the electron, as viewed from the (moving) proton, is as if the proton were fixed and the electron’s mass is reduced to m_r. An expression for m_r is:
\[m_r=\frac{m_e .m_p}{m_e+m_p}\]
\[m_r=0.9995\ m_e\]