Chemlab: Chemistry 6


Spectrum of the Hydrogen Atom

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Chemistry & Background
The line spectra of the elements, like those observed in this experiment, show that electrons in atoms can only exist with discrete, quantized energy values. The state of lowest energy is called the ground electronic state and an electron in this state can absorb but cannot emit energy. Discrete states of higher energy are called excited electronic states. An electron in an excited electronic state can lose energy and change to a state of lower energy. This change of energy state, or energy level, by an electron in an atom is called an electronic transition. The energy lost by the atom, the energy difference between the initial and final states, is emitted as a photon. Since electrons in atoms can exist only with particular, quantized energy values, electronic transitions are also limited to particular energy values. Thus, transitions between electronic energy levels, observed either as emission or absorption of light, occur at discrete energies or wavelengths. In this way, the four visible lines of light emitted by hydrogen atoms in excited electronic states can be used to calculate the differences between energy levels of the electron in a hydrogen atom.

The hydrogen emission spectrum consists of several series of lines, named for their discoverers. A series of emission lines consists of those electronic transitions which all terminate at the same final level. For example, transitions in the Lyman series, which appear in the UV region of the spectrum, all terminate at the ground electronic state of the hydrogen atom. The Paschen, Brackett, and Pfund series of lines are found in the infrared region. In addition there is a series of lines, first discovered by Balmer, in the visible region of the electromagnetic spectrum. The frequencies of the four lines in this series that you will observe can be fit to the Balmer equation:


where n is an integer equal to or greater than 3. Balmer's equation was simply an empirical fit to the observed emission frequencies, without any basis in theory.

The Bohr model of the atom provides a theoretical basis for explaining the line spectra of hydrogen atoms. Based on a planetary model of the atom, Bohr hypothesized that an electron could only exist in quantized energy levels, with the electron orbiting the nucleus at a fixed radius. The allowed quantized energy levels depend on the value of an integer n, called the principal quantum number, which can take any value in the range 1,2,3, ..., ∞. According to Bohr theory, which accurately predicts the energy levels for one-electron atoms like H, He+, Li2+, the energy of an electron in the nth energy level is given by:


where Z is the nuclear charge, -e is the electron charge, me is the mass of the electron, εo is the permittivity of free space, n is the principal quantum number, and h is Planck's constant. Note that the allowed energies are negative numbers and that as n increases, the energy becomes less negative. This means that an electron in a level with n=1 is more tightly bound to the nucleus than an electron in a level with n=2. The zero of energy occurs when n=∞, and for this value of n the allowed Bohr orbit has an infinite radius (this is shown in Eq. 15-7 on p. 539 of Oxtoby, Gillis, and Nachtrieb). Since the zero of energy corresponds to the electron and the nucleus at infinite separation and both at rest, it corresponds to the state of ionization. The energy levels predicted by Bohr theory for the H atom are shown in Figure 1.


Figure 1
Energy Levels in the Bohr Atom
and Electronic Transitions of the Balmer Series

Clearly, electronic transitions between the quantized energy levels of the Bohr atom will give rise to discrete line spectra. For the Balmer series of hydrogen, light is emitted when an electron makes a transition from energy levels with n ≥ 3 to the n = 2 energy level, as shown in Figure 1. The energy of light emitted corresponds to the energy level difference between the final and initial levels (note that Z=1 for H):

ΔE = Ephoton = Efinal - Einitial = hνphoton = hc / λ (3)
(4)
(5)


Since e, me, εo, and h are fundamental constants, this equation expresses the difference between hydrogen atom energy levels in terms of the principal quantum numbers of those levels. This energy level difference corresponds to the energy of the light that is emitted or absorbed when the electron changes its energy. Note that in an emission process, the atom loses energy. Its energy becomes more negative and ΔE for the atom is negative. This is consistent with the above equation since in an emission process nfinal is less than ninitial. In an absorption process, the atom gains energy, nfinal is greater than ninitial and ΔE for the atom is positive. You can also see that the equation has the same form as Balmer's empirical one, with nfinal = 2.

As an example, let's examine the lowest energy line in the Balmer Series, where the electron makes a transition from the n=3 level to the n=2 level. For this case,

= –(2.178 × 10-18 J) ( 1/22 - 1/32) (6)
= –(2.178 × 10-18 J) ( 1/4 - 1/9)
= –3.035 × 10-19 J


As mentioned above, since this is an emission process, ΔE for the atom is negative. Thus, the energy gained by the surroundings, i.e. the energy of the emitted photon, is given by |
ΔE |, and this can be converted to the wavelength of light emitted using the relationship between energy and the wavelength of light, E = h c/λ :

λ = h c / | ΔE | (7)
= 6.5452 × 10-7 m = 654.52 nm

Thus, the lowest energy line (which is the longest wavelength line) in the Balmer series appears in the red portion of the visible spectrum.

Multielectron Atoms and the Effective Nuclear Charge The Bohr model of the atom is incorrect in several important ways, for example, electrons do not move in orbits of fixed radii. However, the more accurate quantum mechanical theory developed by Schrödinger confirms the correctness of the Bohr energy level expression for one-electron atoms: for one-electron atoms the energy of the electron depends only on the value of the principal quantum number n:



For multielectron atoms, quantum mechanics shows that energy levels in such systems are quantized and that the energy of an electronic level depends on both n and the orbital angular momentum quantum number, .

To show how this dependence on arises, we compare the case of the H atom with that of the Na atom. In the H atom, with one electron and one proton, at any instant the electron always experiences the same value of the nuclear charge, namely, +1e. For comparison, consider the ground state Na atom with electron configuration, 1s22s22p63s1. The nuclear charge experienced at any instant by the 3s valence electron depends on its position relative to the nucleus compared to the positions of the 10 core electrons. If the 10 core electrons were always closer to the nucleus than the 3s valence electron, the 3s electron would always experience a nuclear charge of +1e, which is the +11e of the nucleus combined with the –10e charge of the other electrons. If this were the case, the 3s valence electron would be perfectly shielded from the nucleus by the core electrons. However, an examination of the radial probability distribution plots in Figure 2 reveals that there is, at some instant, a significant probability of finding the 3s electron closer to the nucleus than some of the core electrons. At such instants, the 3s electron will experience a nuclear charge that is greater than +1e. At such instants, the 3s electron is said to be imperfectly shielded from the full nuclear charge. Thus, the nuclear charge experienced by the 3s electron varies from instant to instant, and for such an electron we can only define an average or effective nuclear charge, (Zeff)3s.

In the Na atom excited electronic configuration 1s22s22p63p1, the effective nuclear charge experienced by the valence 3p electron will also vary from instant to instant and, in an analogous fashion, we can define an effective nuclear charge, (Zeff)3p, for this electron. A comparison of the radial probability distribution plots in Figure 2 for 3s and 3p electrons, shows that there is a greater probability of finding the 3s electron very close to the nucleus than there is of finding the 3p electron very close to the nucleus. That is, the 3s electron can penetrate to the nucleus, and thereby be closer to the nucleus than some of the core electrons, more frequently than the 3p electron can penetrate to the nucleus. As a result, the 3s electron experiences a nuclear charge greater than +1e more often than does a 3p electron, and (Zeff)3s (Zeff)3p.



Figure 2

At this level of theoretical approximation, the allowed energy levels for a multielectron atom can be expressed as:

(8)

This equation shows that the dependence of E on arises from the dependence of Zeff on .

Since (Zeff)3s > (Zeff)3p, the 3s orbital has a lower energy than the 3p orbital. This is clearly consistent with the arguments presented above; the 3s electron feels a larger effective nuclear charge, is therefore bound more tightly to the nucleus, and thereby has a lower (more negative) energy than the less tightly bound 3p electron.

In the second part of this experiment, you will measure the spectrum of sodium and determine the wavelength of the emission line. From this wavelength, the effective nuclear charges of the 3s and 3p electrons can be calculated.

The sodium emission spectrum has a prominent yellow line, called the sodium D line. This can be observed in the yellow cast of low-energy sodium streetlights. This line arises from the transition of an electron from the excited electronic state in which the valence electron is in a 3p orbital to the ground electronic state in which the valence electron is in a 3s orbital. By measuring the sodium spectrum, you will be able to determine the energy difference between these two electronic states and thereby the energy difference between the 3p and 3s orbitals. The existence of this emission line shows that electrons in the 3s and 3p orbitals are of different energy and that their energy depends on the , as well as the n, quantum number.

To determine the absolute energies of the sodium 3s and 3p orbitals, additional information is required. This is provided by the ground state ionization energy, which is the energy required to remove the 3s valence electron from the ground electronic state of the sodium atom. That is, the ionization energy is the energy of a transition from the 3s level to the n=∞ level. This is shown schematically in Figure 3.



Figure 3
Schematic of Sodium Atom Energy Levels

The ionization energy of gaseous sodium atoms is 496 kJ mol-1, or 8.32 × 10-19 J for a single sodium atom. This value can be used with the wavelength of the sodium D line to determine absolute energy values for the 3s and 3p levels. The wavelength of emitted light corresponds to the difference between the 3s and 3p orbital energies. The ground state ionization energy (IE) is the energy required to transfer the valence electron from the 3s energy level to the n=∞ level, which is defined as the zero of energy. Thus,

IE =Efinal - Einitial =En=∞ - E3s =0 - E3s = -E3s (9)
and ΔED line = Efinal - Einitial = E3s - E3p (10)
therefore E3p = E3s - ΔED line (11)


Recall, that for an emission process, ΔE is negative, since the atom loses energy. Thus, the above equation can be written in the alternate form:

E3p = E3s + | ΔED line | (12)


Once the absolute energy of an orbital and its quantum number n are known, Zeff can be calculated using equation (8). For the sodium portion of this experiment, you will determine the wavelength of the D line, convert this into an energy, ΔED line and calculate the energies and the Zeff values for the 3s and 3p orbitals.

The Meterstick Spectroscope and the Diffraction Equation
In this experiment, a simple spectroscope made from metersticks will be used to observe atomic spectra. The light is supplied by a gas discharge tube, which works like a neon sign. A sample of gas is sealed inside a glass envelope, with electrodes in it. A high voltage is applied across the electrodes and a plasma is formed, with free, accelerated electrons dissociating the hydrogen molecules into excited atoms. These excited atoms emit light, as electrons in excited electronic states make transitions to electronic levels of lower energy. A diffraction grating is used to separate the emitted light into its component wavelengths and a meterstick is used to measure the positions of the emitted lines of light.

A schematic diagram of the meterstick spectroscope is shown in Figure 4. Light from the discharge tube passes through a collimating slit and the incident beam is transmitted through a diffraction grating. A transmission diffraction grating is made by cutting equally spaced parallel grooves (also called rulings) in a glass plate. The incident beam of light is diffracted by the rulings on the grating and emission lines can be viewed along the meterstick, on either side of the incident beam, as indicated by observers 1 and 2 in Figure 4. Emission lines of different wavelengths are diffracted at different angles, θ, and appear at different positions on meterstick a. This is shown by the three different arrows for observer 2. Each of the three emission lines shown on the left side of the slit was diffracted a different angle θ and each has a different wavelength. The diffraction equation discussed in lecture and reproduced below can be used, with the distances from the slit to the observed line and from the slit to the grating to determine the wavelengths of the observed lines of light.


Figure 4
Schematic of the Meterstick Spectroscope, seen from above.
Arrows indicate the path of light.

A derivation of the fundamental diffraction equation required for analysis of the spectral data was given in lecture and is reproduced here. In order for constructive interference to occur at the angle θ, waves from the upper ruling on the diffraction grating must be in phase with waves from the lower ruling, as shown in Figure 5.


Figure 5
A beam of light from the discharge tube is collimated by the slit and strikes the diffraction grating.

Figure 6 shows that this is possible if the path difference TS corresponds to an integral number of wavelengths, λ:
Fig. 6 Close-up of diffraction by grating
Figure 6
Close-up of the diffraction of light by one ruling on the grating. The path difference is TS.

That is,

TS = mλ      where m = 0, 1, 2, 3,... (13)


A consideration of the right-angle triangle RST shows that

sin θ = TS/d (14)


where d is the spacing between centers of adjacent rulings on the diffraction grating. Thus for constructive interference

mλ = d sin θ      where m = 0, 1, 2, 3,... (15)


Here m is an integer called the order of diffraction, d is the spacing between centers of adjacent rulings on the diffraction grating, and θ is the angle, relative to the direction of the incident beam, at which constructive interference occurs.

In this experiment, you will observe the first-order diffraction pattern, so that m is always equal to 1. Thus, the diffraction condition reduces to λ = d sin θ. From this equation it should be clear that for a given value of d (i.e. for a given diffraction grating), the angle θ at which constructive interference occurs will depend on the wavelength, λ, of the emitted radiation. Conversely, this equation shows that the measurement of the angle θ leads directly to a calculation of the wavelength, λ. This diffraction angle, θ, can be determined from the position of the diffracted emission lines on the meterstick, as shown in Figure 7. You may wish to convince yourself of this geometry.

Geometry of the Meterstick Spectroscope Figure 7
Geometry of the Meterstick Spectroscope

For the observation of an emission line at distance a on the meterstick,

θ = arctan (a/b) (16)


To determine the wavelength, λ, of the observed line, this is combined with the first-order diffraction condition to give

λ = d sin θ (17)
= d sin (arctan a/b)


Thus, from measurements of a and b and the spacing (d) between adjacent rulings on the diffraction grating, the wavelength of emission lines can be calculated.

How Does a Fluorescent Light Work?
A fluorescent light operates like a discharge tube, but has been optimized to give diffuse, white light in order to be easier on the eyes. The tube of a fluorescent light bulb contains a low pressure of gas, which emits visible and UV light when a voltage is applied across the tube's electrodes. The inside of the tube is coated with a phosphorescent material that re-emits this light at wavelengths throughout the visible region, making the light from the lamp appear white. In the final part of this experiment, you will compare the spectrum of a fluorescent light to that of several elements, to determine what gas is inside the fluorescent tube.
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