Physics 321 

Prof. Dale E. Gary
NJIT 
Classification of Spectra,
and HertzsprungRussell Diagrams
Spectra from the Hydrogen Atom
The Bohr model of the hydrogen atom gave us the following results:Excitation, Deexcitation, Ionization and RecombinationOne way to see why the angular momentum should be quantized is to consider the electron as a wave. As de Broglie showed, particles exhibit wave characteristics with wavelength given by
 The angular momentum, l, was postulated to be quantized according to
l = mvr = nh/2p. where n is the principle quantum number.
 The radii of allowed electron orbits were also quantized according
r = a_{o}n^{2} = n^{2}h^{2}/(4p^{2}ke^{2}m)
where a_{o}= 5.29 x 10^{11} m is the Bohr radius. The energies of the orbits were likewise quantized according to
E_{n} = R' [1/n^{2}]
where R' = 13.6 eV is a constant. The electronvolt (eV) is a unit of energy (the energy an electron would gain after acceleration through a potential of 1 volt). 1 eV = 1.602 x 10^{19} J The difference between energy levels (orbits) corresponds to a wavelength
1/l_{ab} = n_{ab}/c = (E_{b } E_{a})/ch=R [1/n_{b}^{2}1/n_{a}^{2}]
where R = 10.96776 mm^{1} is the Rydberg constant.l = h/p = h/mvWhat if the lowest orbit corresponded to one wavelength of such an electron wave? The circumference of the orbit would correspond to a wavelength, so2pr = l = h/mv => l = mvr = h/2p.The next higher orbit would correspond to two electron wavelengths, and the nth orbit to n electron wavelengths, so in this way we naturally find l = nh/2p. If the orbit corresponded to a noninteger multiple of the electron wavelength, the electron wave would overlap and interfere with itself.Here is a scale model of the orbits of the hydrogen atom, with the radii getting further apart according to n^{2}:
However, despite the orbits getting ever farther apart in space, they get closer together in energy according to 1/n^{2}. Any energy level diagram is shown below:
When an electron jumps from a lower energy level to a higher energy level (smaller n to larger n), the atom is said to be excited, and the process is called excitation. Energy must be added to the atom in order for such a jump to occur. There are two ways for an atom to be excited:Heisenberg Uncertainty PrincipleThere are also two reverse processes by which the electron can jump from the higher energy level to a lower one, in the process of deexcitation. These are:
 If the atom suffers a collision that causes an electron to jump from the lower to higher energy state it is called collisional excitation.
 If the electron captures a photon of energy hn corresponding to the energy difference between the two energy levels, and so jumps from the lower to higher energy state, it is called radiative excitation.
In addition to electrons jumping up and down between energy levels in the atom, it is possible for an electron to jump completely away from the atom (ionization), or a free electron to be captured into an energy level (recombination).
 If the atom suffers a collision that causes an electron to jump from a higher to lower energy state it is called collisional deexcitation.
 If the atom jumps from a higher to lower energy state and emits a photon of energy hn corresponding to the energy difference between the two energy levels, it is called radiative deexcitation. Since energy is released during such a transition, the transition is spontaneous, meaning it occurs without any outside action.
We noted last time the relationships:DxDp ~ h/2pbut did not have time to show some effects that come from them. The n = 2 level of hydrogen lives for a very short time, only about 10^{8} s. That is, an electron in the first excited state will, after about 10^{8} s, spontaneously emit a photon of energy 10.2 eV and go back to the ground state (emitting a Lyman a photon). The second relation, above, means that the photons emitted have an energy uncertainty of
DED t ~ h/2pDE ~ h/2pD t = 1.05 x 10^{26} J => Dl ~ l^{2}/2pD t c = 0.014 mA
This uncertainty in the wavelength shows up as a broadening of the spectral line (natural broadening).Spectral ClassificationSimilarly, we can ask what is the uncertainty in speed that is required of an electron that is confined in a hydrogen atom (in the ground state orbit, with Bohr radius a_{o}= 5.29 x 10^{11} m).
Example: Imagine an electron comfined within a region of space the size of a hydrogen atom. What is the minimum speed and energy of the electron, estimated using the Heisenberg uncertainty principle? We know Dx ~ a_{o}= 5.29 x 10^{11 }m soDp ~ h/2pa_{o} = 6.62 x 10^{34} J s/2p(5.29 x 10^{11 }m) = 1.99 x 10^{24 }kg m/s.This is the uncertainty in the momentum, which is about equal to the minimum momentum p_{min} that the electron has in order to be confined within the atom. Now the momentum is just p = mv, so the minimum velocity isv_{min} = p_{min}/m_{e} = 1.99 x 10^{24} kg m/s / 9.1 x 10^{31} kg = 2.19 x 10^{6} m/s.The corresponding minimum kinetic energy isK_{min} = 1/2m_{e}v^{2}_{min}= 2.18 x 10^{18} Jwhich may seem like a small energy, but when converted to electron volts, this isK_{min} = 2.18 x 10^{18} / 1.602 x 10^{19} J/eV = 13.6 eV!This is exactly the energy of an electron in the ground state of the hydrogen atom! Later we will see that this subtle quantum effect is responsible for supporting the tremendous gravitational weight inside a white dwarf star.
Adapted from data in the electronic version of "A Library of Stellar
Spectra,"
by Jacoby G.H., Hunter D.A., Christian C.A. Astrophys. J.
Suppl. Ser., 56, 257 (1984).
The spectra were then reordered
according to temperature using the
continuum spectrum (blackbody curve) as a guide (Annie Jump Cannon), to
arrive at a new ordering:
...B7, B8, B9, A0, A1, A2,
A3, ..., A7, A8, A9, F0, F1... etc.
(early A)
(late A)
These questions can be answered by appealing to statistical mechanics. Imagine firing a beam of particles, all with the same energy, into a trap and allowing them to come into equilibrium by colliding elastically. Due to standard probability and statistics (akin to cointossing), they will wind up with a gaussian distribution of speeds (a bell curve). Any distribution of particles in equilibrium will have such a curve, which can be parametrized by a single number, the temperature. Due to the 3D nature of velocity, the gaussian when expressed in terms of speed becomes:Boltzmann Equationn_{v} dv = n (m/2pkT)^{3/2}exp(1/2 mv^{2}/kT )(4pv^{2}) dvwhere the important dependence is the argument of the exponential, which is just the ratio of the kinetic energy (1/2 mv^{2}) to thermal energy (kT ). This expression is the number of particles per unit volume having speeds between v and v + dv.Important relationships:
 Peak of distribution (most probable speed) is when 1/2 mv^{2} = kT, or
v_{mp} = [2kT/m]^{1/2}.
 Mean speed is somewhat higher (due to asymmetry in the curve)
v_{mean} = [3kT/m]^{1/2}.
As atoms collide, their electrons can be knocked up to the next higher energy level if the colliding atoms have enough energy (collisional excitation). The electron can even be knocked entirely away from the atom (ionization). Again, looking at the problem from a statistical standpoint, the probability of the atom's being in one energy state, s_{a}, isSaha EquationP(s_{a}) ~ exp(E_{a}/kT )and the probability for state s_{b} isP(s_{b}) ~ exp(E_{b}/kT )where E_{a} and E_{b} are the energies of the two states (e.g. = 13.6 eV for the ground state of the hydrogen atom). The ratio of these probabilities is then
P(s_{b}) exp(E_{b}/kT ) = P(s_{a}) exp(E_{a}/kT ) However, more than one state in an atom can have the same energy (i.e. they can be degenerate), e.g. in He I (neutral helium) two electrons in the ground state have n = 1, but the two electrons spin in opposite directions (m_{s} = +1/2, 1/2). We define the degeneracy (the number of states with the same energy E_{a}) as g_{a}, which is called the statistical weight of state a. The above expression then becomes
P(s_{b}) g_{b} exp(E_{b}/kT ) = P(s_{a}) g_{a} exp(E_{a}/kT ) For a large number of atoms, the ratio of probabilities must be the same as the ratio of numbers of atoms in the two energy levels, e.g.
N_{b } g_{b } = exp([E_{b}E_{a}]/kT ) N_{a } g_{a} This is the Boltzmann Equation. It turns out that for hydrogen, the statistical weights are given by g_{n}= 2n^{2}. The Boltzmann Equation answers the first question we had, namely, how does the population of levels depend on T. However, if we want to populate level 2 to make strong Hbalmer lines, what temperature do we need, say, to make as many level 2 atoms as there are level 1 atoms? The equation becomes
N_{2 } 8 = exp([E_{2}E_{1}]/kT ) = 4 exp([10.2 eV+13.6 eV]/kT ) = 1 N_{1 } 2 but when we solve for T we find T ~ 85,000 K! Much higher than the observed 9,500 K. Clearly there is another important effect that we did not include. That other effect is ionization.
To get a hydrogen atom's electron into level 2, it required a collision with at least 10.2 eV of energy. However, once it is in level 2, it requires only 3.4 eV more energy to knock it completely away from the atom, so the first excited state is a very precarious place for an electron! The actual number of atoms in level 2 at any one time is a balance between collisions to get it there in the first place, and collisions to ionize the atom.Partition FunctionLet c_{i} be the ionization potential of the ith ionization stage. This is the amount of energy needed to remove an electron from its lowest energy state. For the ionization stage H I, for example, it would be 13.6 eV. For He I it would be roughly the same, but for He II it would be four times larger (why?). You might think that the ratio of atoms in one stage relative to another would be
N_{i+1 } ~ exp( c_{i }/ kT ) N_{i } which is nearly right, but not all of the atoms are in this lowest state, so we have to add up statistically all of the ways an atom in various energy states can be stripped of an electron.
_{ } infinity Z_{i } = S g_{1} exp([E_{b}E_{a}]/kT ) _{ } _{ 1=1} where the sum is over the distribution of energy states of an atom at temperature T from the Boltzmann Equation.
Then
Combining the Boltzmann and Saha Equations
N_{i+1 } Z_{i+j } ~ exp( c_{i }/ kT ) N_{i } Z_{i} which leads to the Saha Equation
N_{i+1 } 2Z_{i+j } ( 2pm_{e}kT ) = exp( c_{i }/ kT ) N_{i } n_{e}Z_{i} h^{2} Example: Saha Equation for H between 5,000 and 25,000 K.
Partition function for H II is Z_{II}= 1, since there is no degeneracy for a bare proton. For H I the summation can be done over just two terms (why?) which givesZ_{II}= 2 + 8 exp([(10.2 eV)(13.6 eV)]/kT ) = 2 + 8 exp(3.4 eV/kT )Now, writing the thermal energy in eV, we have the range ~ 0.5 to 2.5 eV, so the exponential term ranges from about 8(0.001) to 8(0.25), soZ_{II}= 2 to 4over the temperature range. What we really want is not N_{II}/N_{I}as given by the Saha equation, but a related quantity N_{II}/N_{total }= N_{II}/(N_{I}+N_{II}) = N_{II}/N_{I }/ (1+N_{II}/N_{I}) which is plotted in Figure 813 in the text (curve labelled log N_{+}/N_{o}). You can see that almost all of the H will be ionized by about 10,000 K. This is what limits the strength of the Hbalmer lines above 9,500 K.
We can find the variation of the strength of the Halpha line with temperature by combining the Boltzmann and Saha equations as follows:
N_{2} N_{2} N_{I} = N_{total} N_{1}+N_{2 } N_{total} or
N_{2} N_{2}/N_{1} N_{I} N_{2}/N_{1} 1 = = N_{total} 1+N_{2}/N_{1} N_{I}+N_{II} 1+N_{2}/N_{1} 1+N_{II}/N_{I} The final expression is written entirely in terms of the quantities provided by the Boltzmann and Saha equations. The full dependence is plotted in Figure below, and in Fig. 813 of the text. Here we see very nice confirmation of the observed strength of the hydrogen lines, since the curve peaks right near 10,000 K as required to match the observations.