In a previous set of notes, we looked at the planetary model of the atom as proposed by the Danish physicist Niels Bohr. While that model was sufficient to understand why certain atoms form ions of one particular charge but no other charges, the model still has several shortcomings that require quantum theory to resolve.
The Bohr model pictures the electrons as orbiting around the atomic nucleus, much as planets orbit around the sun, but we know today that electrons do not have a precise trajectory that we can localize. Rather, we must speak in terms of the probability of finding an electron in a particular region of space. The "orbits" that we draw in the Bohr model correspond to the average distance of an electron from the atomic nucleus. An electron that resides in the second shell is, on average, more distant from the atomic nucleus than one that resides in the first shell.
Even though we know electrons don't move in precise circular orbits, the concept of electron shells is retained in quantum theory, and the quantum number n is used to designate the shells. It is with this idea in mind that we refer to the first orbit in the Bohr model as the n=1 shell, the second orbit in the Bohr model as the n=2 shell, and so on. But electrons come much closer to having precise energies than precise orbits. For the hydrogen atom, which has only a single electron, we can actually calculate these energies from the principles of quantum mechanics. The energy of an electron in the hydrogen atom depends on the shell it is in, and is given by the formula
R_{H}
E_{n} =  
(Equation 1)
n^{2}
where R_{H}, known as the Rydberg constant, has the value 2.1798741 x 10^{18} J and n designates the shell. Notice that the formula has a negative sign in front of it. This results from the position of the zero energy reference point. When the electron is an infinite distance from the nucleus, the system is taken to have zero energy. Of course, "infinite distance" is unattainable, but in practice, "infinite distance" is when the electron and the atomic nucleus are so far apart that they do not interact with one another. When the electron is brought closer to the nucleus, the positive nucleus attracts the negative electron. Since attractive interaction lowers the system energy, and we took the zero of energy to be "infinite" separation, the energy must be negative on a relative scale. This is not saying that the system has negative energy, just that its energy is lower than that at the zero energy reference point. You should be familiar with the idea of relative energies from your previous study of thermochemistry. When we looked at enthalpy changes (DH) for chemical reactions, we noted that we could not determine the absolute amount of energy locked up in a chemical system. We had to arbitrarily set the elements to zero enthalpy. The enthalpy of formation of a compound was then determined as the change in enthalpy that occurs when one mole of the compound forms directly from the elements.
In the above equation giving electron energies in the hydrogen atom, note that the quantum number n is restricted to positive integers. That is,
n = 1, 2, 3, . . . .
Zero, negative, and fractional values are not allowed. The electron has its lowest energy when it is in the shell closest to the nucleus, which is the n=1 shell. For any atom, the electron configuration of lowest possible energy is called the ground state. For the hydrogen atom, its one single electron must be in the n=1 shell to be in the ground state. If the atom absorbs energy (for example, the system is intensely heated, or subjected to an electric discharge), the electron may be promoted to a state of higher energy. When this happens, the atom is said to be in an excited state.
One of the most significant tenants of quantum mechanics is quantization of energy, that is, the idea that only certain values of energy are allowed. This is contrary to our normal experience. For example, if we throw a baseball, we perceive that we can throw the ball with any degree of energy that our muscles are capable of imparting to it. For macroscopic objects  such as the baseball  the allowed energy levels are so closely spaced that we do not notice the quantization of energy. But for objects on a subatomic scale, the quantum mechanical properties become significant. We see quantization in the energy level formula, because the value of n is restricted to positive whole numbers. Any energy that would imply a fractional value of n is not allowed. To 4 significant figures, the hydrogen atom electron energy for shells 1 through 6 are calculated as follows:
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{1} =  
=   = 2.180 x
10^{18} J
1^{2}
1
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{2} =  
=   = 5.450 x
10^{19} J
2^{2}
4
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{3} =  
=   = 2.422 x
10^{19} J
3^{2}
9
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{4} =  
=   = 1.363 x
10^{19} J
4^{2}
16
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{5} =  
=   = 8.720 x
10^{20} J
5^{2}
25
2.180 x 10^{18}
J
2.180 x 10^{18} J
E_{6} =  
=   = 6.056 x
10^{20} J
6^{2}
36
If we plot each of these allowed energies as a horizontal line on a graph, we get the following result:
Figure 1. Allowed energy levels for an electron in a hydrogen atom. The top line in the diagram corresponds to the zero of energy, making all the other energy levels negative in a relative sense. The quantum number n is equal to infinity at the zero energy line. The quantum states n=1 through n=6 are shown in the diagram, though only the first four states are labeled, due to space limitations. All quantum energies above n=6 are located between the n=6 line and the zero energy line.
Notice in Figure 1 that the allowed energies get closer and closer together as the energies become higher and higher. This is because qunatization of energy only occurs in "bound states", that is, states in which the electron is under the influence of the attractive pull toward the atomic nucleus. If the electron was totally free, there would be no quantization of energy. As the electron is taken further and further away from the nucleus, the strength of the attractive force decreases. That is, as the electron gets further away from the nucleus, it becomes more like a free particle. As a result, the quantization of energy becomes less pronounced, and the energy levels eventually merge into a continuum.
At room temperature, we expect to find the vast majority of hydrogen atoms in the ground state (that is, with the electron in the n=1 state), but if we impart the right amount of energy to the atom (for example, by generating an electric discharge through a tube of low pressure hydrogen gas) we can promote the electrons to higher energy states. In this process, the hydrogen atom can only absorb energies in amounts that correspond to differences between two allowed electron energy levels. For example, if a hydrogen atom's electron currently resides in the n=1 shell, the amount of energy the atom must absorb to promote the electron to the n=2 shell is calculated as follows:
DE_{atom} = E_{2}  E_{1} = (5.450 x 10^{19} J)  (2.180 x 10^{18}J) = 1.635 x 10^{18} J
Notice that the atom's change in energy is positive, because the atom has absorbed energy in order to promote the electron from the n=1 state to the n=2 state. We follow the same sign convention here as we did in thermochemistry  the energy change is positive when energy is absorbed and negative when energy is given off.
Now consider a hydrogen atom that has previously absorbed energy to promote its electron to the n=3 state. If this electron deexcites from the n=3 state to the n=2 state (symbolized by the notation 3>2) the atom's change in energy is
DE_{atom} = E_{2}  E_{3} = (5.450 x 10^{19} J)  (2.422 x 10^{19} J) = 3.028 x 10^{19} J
In this case the atom has a negative change in energy, because the electron must lose energy in order to drop from the n=3 state to the n=2 state.
While the atom can absorb the energy needed to excite the electron either as heat or light, the energy is always returned as light when the electron deexcites. Therefore, we need to understand something about the nature of light. When someone mentions light, they usually mean the light we see by, but this is actually just a small slice of what is more generally known as the electromagnetic spectrum.
Electromagnetic radiation is an electromagnetic disturbance that can travel through space without needing a medium to conduct it. Sunlight reaches earth after traveling through 92 million miles of vacuum. Electromagnetic radiation travels in the form of a wave. If you have ever tossed a stone into a quiet pond and and observed the ripples that travel outward from the point of impact, you are already familiar with wave phenomena.
We can characterize a wave in two ways: by its frequency and by its wavelength. Going back to the stone in the pond example, suppose you have a cork floating on the water a short distance from where the stone impacts the water's surface. As the waves travel across the place where the cork is floating, the cork encounters a series of peaks and valleys in the water level that cause it to bob up and down. If we count the number of times the cork bobs up and down in a given period of time, we get the frequency. To be more precise, the frequency is the number of "cycles" that pass by a given point per unit time. A cycle is complete when the waveform begins to repeat itself. If we begin our observation as the crest of a wave passes by, the arrival of the next crest is the end of the first cycle and the beginning of the next one. We can also characterize a wave by its wavelength. This is the distance between identical points in two consecutive cycles. We often measure this as the distance between two consecutive crests, though any point on the waveform would do.
If we multiply the wavelength by the frequency, we get the speed of the wave. This is logical if you think about it. The wavelength is how long each wave is, and the frequency is the number of waves that pass by in a chosen unit of time. If a certain waveform is 5 cm long and 10 waves pass by in one second, then the wavefront must move 50 cm in that one second interval.
5 cm 10
waves 50 cm
 x  = 
wave
sec
sec
We often consider the unit "wave" or "cycle" to be understood and just report the wavelength as a length (not a length per wave) and the frequency as a reciprocal time (not waves per time). In this case, the calculation is written as follows:
5 cm x 10 sec^{1} = 50 cm sec^{1}
In the above, a power of 1 has been used to indicate that the unit sec appears in the denominator of a fraction.
We usually use the Greek letter lambda (l) to symbolize the wavelength and the Greek letter nu (n, pronounced "new") to symbolize the frequency. When the waveform being considered electromagnetic radiation traveling in a vacuum, the speed of the wave is the well known "speed of light", symbolized by the letter c. The currently accepted value is as follows:
c = 2.99792458 x 10^{8} m ^{.} s^{1}
The speed of electromagnetic radiation in a medium (such as air) will be a little lower than that in a vacuum, but we will normally ignore the small difference and treat it as if the speed is still the speed of light in a vacuum. Since the wavelength and frequency multiply out to the speed of the wave, we have the following equation for wavelength and frequency of electromagnetic radiation:
l ^{.} n = c (Equation 2)
This means that wavelength and frequency are not independent. Once either of these has been specified, the other is automatically set. We can solve Equation 2 for wavelength and frequency to obtain the following:
c
l =
 (Equation 2a)
n
c
n =
 (Equation 2b)
l
Equation 2a allows us to convert frequency to wavelength and Equation 2b allows us to convert wavelength to frequency.
In addition to thinking of electromagnetic radiation as a wave, we can also think of it as a stream of flowing particles. These particles are called photons. The photons carry the energy of the electromagnetic radiation. The photon energy is related to the frequency. The higher the frequency, the higher the photon energy. The equation relating these two quantities is
E_{photon} = h ^{.} n (Equation 3)
where h = 6.6260755 x 10^{34} J ^{.} s
The quantity symbolized above by h is known as Plank's constant. For electromagnetic radiation of known frequency, Equation 3 allows us to calculate the energy of its photons. If we already know the photon energy and want to calculate the frequency that corresponds to it, we can solve Equation 3 for the frequency to obtain
E_{photon}
n =
 (Equation
3a)
h
If we substitute Equation 2b into Equation 3, we obtain the following equation, which relates photon energy to wavelength:
h ^{.} c
E_{photon} =
 (Equation 4)
l
Equation 4 allows us to calculate the photon energy for electromagnetic radiation of a given wavelength. If we already know the photon energy and want to find the wavelength that corresponds to it, we can solve Equation 4 for the wavelength to obtain
h ^{.} c
l =

(Equation 4a)
E_{photon}
From the above considerations, we see that wavelength, frequency, and photon energy are all related. If any one of these three quantities is known, the other two are automatically set, because the equations presented above must be satisfied.
We now turn our attention to the emission spectrum of hydrogen. As noted previously, when an electron in a hydrogen atom deexcites, energy is given off in the form of light (electromagnetic radiation). Each electron deexcitation will give off a photon carrying an amount of energy equal to that which the electron lost in the transition. (A change of an electron from one energy state to another is called a transition) That is,
E_{photon} =  DE_{atom}  (Equation 5)
In Equation 5, absolute value signs appear around DE_{atom} because the energy change for the atom is negative when an electron undergoes a deexcitation, but photon energies are always positive. We normally calculate a change (D) in a quantity by subtracting the initial value from the final value, but to ensure we get a positive answer  as required by Equation 5  we always subtract the lower electron energy from the higher electron energy, even if the electron ends in the lower state (deexcitation). That is,
 DE_{atom}  = E_{photon} = E_{h}  E_{l} (Equation 5a)
In Equation 5a, E_{h} refers to the higher quantum state and E_{l} refers to the lower quantum state.
Earlier in these notes, we considered the change in energy that a hydrogen atom undergoes as its electron undergoes a transition from the n=3 state to the n=2 state, that is, a 3>2 transition:
DE_{atom} = E_{2}  E_{3} = (5.450 x 10^{19} J)  (2.422 x 10^{19} J) = 3.028 x 10^{19} J
Since the atom lost 3.028 x 10^{19} J, that's the amount of energy that must be present in the emitted photon. But in the photon, that amount of energy must be expressed as a positive quantity. Applying Equation 5a, we have
E_{photon} =  DE_{atom}  = E_{3}  E_{2} = (2.422 x 10^{19} J)  (5.450 x 10^{19} J)
= 3.028 x 10^{19} J
Now that we have the photon energy, we can go on to calculate the frequency and wavelength. Using Equation 3a, we can calculate the frequency as follows:
E_{photon}
3.028 x 10^{19} J
n =
 = 
= 4.570 x 10^{14} s^{1}
h
6.626 x 10^{34} J ^{.} s
And using Equation 4a, we can calculate the wavelength:
h ^{.} c
(6.626 x 10^{34} J ^{.} s) (2.998 x 10^{8} m ^{.}
s^{1})
l =
 = 
E_{photon}
3.028 x 10^{19} J
= 6.560 x 10^{7} m
Very small wavelenths like the above are usually expressed in a smaller unit, such as micrometers (mm) or nanometers (nm).
1 nm
6.560 x 10^{7} m x 
= 656.0 nm
1 x 10^{9} m
While the above examples were for a 3>2 electron transition, the same techniques could be used for any electron transition of interest. However, we can also derive generic formulas for the photon energy, frequency, and wavelength of any electron transition in the hydrogen atom. These formulas will include variables for the quantum numbers of the higher and lower quantum states involved in the electron transition.
To derive a generic photon energy formula, we apply Equation 1 to the quantum states E_{h} and E_{l} in Equation 5a. For the quantum state having the energy E_{h}, we refer to the quantum number as n_{h}, and for the quantum state having the energy E_{l}, we refer to the quantum number as n_{l}. We can then write the following equation:
^{} R_{H}
^{} ^{}
R_{H} ^{}
E_{photon} = 
    
  
_{} n^{2}_{h}
_{} _{}
n^{2}_{l} _{}
^{} R_{H}
^{} ^{}
R_{H} ^{}
=
  ^{ }
   
_{} n^{2}_{l}
_{} _{}
n^{2}_{h} _{}
^{}
1
1 ^{}
= R_{H} 
 

 (Equation 6)
_{} n^{2}_{l}
n^{2}_{h} _{ }
The sequence of rearrangements seen in Equation 6 result from recognizing that subtracting a negative quantity changes it to positive and that the factor of R_{H} appears in both terms and can be factored out. This equation allows us to plug in any two quantum states n_{l} and n_{h} and calculate the energy of the emitted photon when a hydrogen atom electron undergoes a transition from the higher state (n_{h}) to the lower state (n_{l}). For example, if we apply Equation 6 to the 3>2 transition we considered previously, we have
^{}
1
1 ^{}
E_{photon} = 2.180 x 10^{18}
J  
 

_{} 2^{2}
3^{2} _{ }
^{}
1
1 ^{}
= 2.180 x 10^{18} J
  
 
_{}
4
9 _{ }
^{}
9
4 ^{}
= 2.180 x 10^{18} J
  
 
_{}
36
36 _{ }
^{} 5 ^{}
= 2.180 x 10^{18} J 
 
_{} 36 _{}
= 3.028 x 10^{19} J
This is the same result we obtained previously, thereby confirming the formula is valid.
Since photon energy, frequency, and wavelength are all related, we can also derive equations for the frequency and wavelength associated with hydrogen atom electron transitions. If we substitute Equation 3 into Equation 6, we get
^{}
1
1 ^{}
h ^{.} n
= R_{H} 
 


(Equation 7)
_{} n^{2}_{l}
n^{2}_{h} _{ }
Solving this for the frequency, n, we have
^{} R_{H} ^{}
^{}
1
1 ^{}
n = 
  
 

 (Equation 7a)
_{} h _{}
_{} n^{2}_{l}
n^{2}_{h} _{ }
If we apply Equation 7a to the 3>2 electron transition we have been considering, we get
^{} 2.180 x 10^{18}
J ^{} ^{}
1
1 ^{}
n = 
 
  
 
_{} 6.626 x 10^{34} J ^{.}
s _{} _{}
2^{2}
3^{2} _{ }
^{} 2.180 x 10^{18}
J ^{} ^{}
1
1 ^{}
=
  
  
 
_{} 6.626 x 10^{34} J ^{.}
s _{} _{}
4
9 _{ }
^{} 2.180 x 10^{18}
J ^{} ^{}
9
4 ^{}
=
  
  
 
_{} 6.626 x 10^{34} J ^{.}
s _{} _{}
36
36 _{ }
^{} 2.180 x 10^{18}
J ^{} ^{}
5 ^{}
=
  
  
_{} 6.626 x 10^{34} J ^{.}
s _{} _{}
36 _{ }
^{} 5 ^{}
= 3.290 x 10^{15} s^{1}
  
_{} 36 _{ }
= 4.569 x 10^{14} s^{1}
This is certainly within round off error of the previous result of 4.570 x 10^{14} s^{1}. Finally, we can derive an equation for the wavelength associated with a hydrogen atom electron transition by substituting Equation 2b into Equation 7a. This gives the following result:
c
^{} R_{H} ^{}
^{}
1
1 ^{}
 = 
  
 

 (Equation 8)
l
_{} h _{}
_{} n^{2}_{l}
n^{2}_{h} _{ }
Dividing both sides of Equation 8 by the speed of light, c, gives
1
^{} R_{H} ^{}
^{}
1
1 ^{}
 = 
  
 

 (Equation 8a)
l
_{} h c _{}
_{} n^{2}_{l}
n^{2}_{h} _{ }
Equation 8a gives the reciprocal of the wavelength rather than the actual wavelength. Solving Equation 8a for the wavelength, l, gives a rather awkward looking result. Furthermore, it is a simple matter to push the reciprocal key (labeled 1/x or x^{1}) on a calculator to change the reciprocal wavelength 1/l to the actual wavelength l. Therefore, we shall take Equation 8a to be our final result for electron transition wavelength, and will not attempt to reexpress the equation to give wavelength directly.
If we apply Equation 8a to the 3>2 electron transition we have been considering, we have the following:
1
^{}
2.180 x 10^{18}
J
^{} ^{}
1
1 ^{}
 = 
 
  
 
l
_{} (6.626 x 10^{34} J ^{.} s)
(2.998 x 10^{8} m ^{.} s^{1}) _{}
_{} 2^{2}
3^{2} _{ }
^{}
2.180 x 10^{18}
J
^{} ^{}
1
1 ^{}
= 
 
  
 
_{} (6.626 x 10^{34} J ^{.} s)
(2.998 x 10^{8} m ^{.} s^{1}) _{}
_{}
4
9 _{ }
^{}
2.180 x 10^{18}
J
^{} ^{}
9
4 ^{}
= 
 
  
 
_{} (6.626 x 10^{34} J ^{.} s)
(2.998 x 10^{8} m ^{.} s^{1}) _{}
_{}
36
36 _{ }
^{}
2.180 x 10^{18}
J
^{} ^{}
5 ^{}
= 
 
  
_{} (6.626 x 10^{34} J ^{.} s)
(2.998 x 10^{8} m ^{.} s^{1}) _{}
_{} 36 _{ }
^{} 5 ^{}
=
(1.097 x 10^{7} m^{1})
 
 = 1.524 x 10^{6} m^{1}
_{} 36 _{ }
l = (1/l)^{1} = (1.524 x 10^{6} m^{1})^{1} = 6.562 x 10^{7} m
This is within round off error of the previously obtained result of 6.560 x 10^{7} m.
Notice that in equations 6, 7a, and 8a, a constant multiplies the difference of the reciprocals of the squares of the quantum numbers. As you apply these formulas to different problems, the only thing that changes is the particular quantum numbers involved. If you evaluate the constants in these equations (it would be a good idea to use the full unrounded values of R_{H}, h, and c), you can store the results in your calculator's memories and recall them quickly during problem solving. You can greatly increase your problem solving speed in this way.