6 h spectrum

Spectrum of theHydrogen Atom

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Chemistry & BackgroundThe line spectra of the elements, like those observed in thisexperiment, show that electrons in atoms can only exist withdiscrete, quantized energy values. The state of lowest energyis called the ground electronic state and an electron in thisstate can absorb but cannot emit energy. Discrete states ofhigher energy are called excited electronic states. An electronin an excited electronic state can lose energy and change to astate of lower energy. This change of energy state, or energylevel, by an electron in an atom is called an electronictransition. The energy lost by the atom, the energy differencebetween 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 alsolimited to particular energy values. Thus, transitions betweenelectronic energy levels, observed either as emission orabsorption of light, occur at discrete energies or wavelengths.In this way, the four visible lines of light emitted by hydrogenatoms in excited electronic states can be used to calculate thedifferences between energy levels of the electron in ahydrogen atom.

The hydrogen emission spectrum consists of several series oflines, named for their discoverers. A series of emission linesconsists of those electronic transitions which all terminate atthe same final level. For example, transitions in the Lymanseries, which appear in the UV region of the spectrum, allterminate at the ground electronic state of the hydrogenatom. The Paschen, Brackett, and Pfund series of lines arefound in the infrared region. In addition there is a series oflines, first discovered by Balmer, in the visible region of theelectromagnetic spectrum. The frequencies of the four linesin this series that you will observe can be fit to the Balmerequation:

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

The Bohr model of the atom provides a theoretical basis forexplaining the line spectra of hydrogen atoms. Based on aplanetary model of the atom, Bohr hypothesized that anelectron could only exist in quantized energy levels, with theelectron orbiting the nucleus at a fixed radius. The allowedquantized energy levels depend on the value of an integer n,called the principal quantum number, which can take any

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value in the range 1,2,3, ..., . According to Bohr theory,which accurately predicts the energy levels for one-electronatoms like H, He+, Li2+, the energy of an electron in the nthenergy level is given by:

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

Figure 1 Energy Levels in the Bohr Atom

and Electronic Transitions of the BalmerSeries

Clearly, electronic transitions between the quantized energylevels of the Bohr atom will give rise to discrete line spectra.


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For the Balmer series of hydrogen, light is emitted when anelectron makes a transition from energy levels with n 3 tothe n = 2 energy level, as shown in Figure 1. The energy oflight emitted corresponds to the energy level differencebetween the final and initial levels (note that Z=1 for H):

E = Ephoton = Efinal - Einitial = hphoton = hc / (3)(4)

(5)

Since e, me, o, and h are fundamental constants, thisequation expresses the difference between hydrogen atomenergy levels in terms of the principal quantum numbers ofthose levels. This energy level difference corresponds to theenergy of the light that is emitted or absorbed when theelectron changes its energy. Note that in an emission process,the atom loses energy. Its energy becomes more negative andE for the atom is negative. This is consistent with the aboveequation since in an emission process nfinal is less thanninitial. In an absorption process, the atom gains energy, nfinalis greater than ninitial and E for the atom is positive. Youcan also see that the equation has the same form as Balmer'sempirical one, with nfinal = 2.

As an example, let's examine the lowest energy line in theBalmer Series, where the electron makes a transition from then=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, E forthe atom is negative. Thus, the energy gained by thesurroundings, i.e. the energy of the emitted photon, is givenby |E |, and this can be converted to the wavelength of lightemitted using the relationship between energy and thewavelength of light, E = h c/ :

= h c / | E | (7)


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= 6.5452 10-7 m = 654.52 nm

Thus, the lowest energy line (which is the longestwavelength line) in the Balmer series appears in the redportion of the visible spectrum.

Multielectron Atoms and the Effective Nuclear ChargeThe Bohr model of the atom is incorrect in several importantways, for example, electrons do not move in orbits of fixedradii. However, the more accurate quantum mechanicaltheory developed by Schrdinger confirms the correctness ofthe Bohr energy level expression for one-electron atoms: forone-electron atoms the energy of the electron depends onlyon the value of the principal quantum number n:

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

To show how this dependence on arises, we COMPAREthe case of the H atom with that of the Na atom. In the Hatom, with one electron and one proton, at any instant theelectron always experiences the same value of the nuclearcharge, namely, +1e. For comparison, consider the groundstate Na atom with electron configuration, 1s22s22p63s1. Thenuclear charge experienced at any instant by the 3s valenceelectron depends on its position relative to the nucleusCOMPARED to the positions of the 10 core electrons. Ifthe 10 core electrons were always closer to the nucleus thanthe 3s valence electron, the 3s electron would alwaysexperience a nuclear charge of +1e, which is the +11e of thenucleus combined with the 10e charge of the otherelectrons. If this were the case, the 3s valence electron wouldbe perfectly shielded from the nucleus by the core electrons.However, an examination of the radial probabilitydistribution plots in Figure 2 reveals that there is, at someinstant, a significant probability of finding the 3s electroncloser to the nucleus than some of the core electrons. At suchinstants, the 3s electron will experience a nuclear charge thatis greater than +1e. At such instants, the 3s electron is saidto be imperfectly shielded from the full nuclear charge. Thus,the nuclear charge experienced by the 3s electron varies frominstant to instant, and for such an electron we can only definean average or effective nuclear charge, (Zeff)3s.

In the Na atom excited electronic configuration


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1s22s22p63p1, the effective nuclear charge experienced bythe valence 3p electron will also vary from instant to instantand, in an ANALOGOUS fashion, we can define aneffective nuclear charge, (Zeff)3p, for this electron. Acomparison of the radial probability distribution plots inFigure 2 for 3s and 3p electrons, shows that there is a greaterprobability of finding the 3s electron very close to thenucleus than there is of finding the 3p electron very close tothe nucleus. That is, the 3s electron can penetrate to thenucleus, and thereby be closer to the nucleus than some ofthe core electrons, more frequently than the 3p electron canpenetrate to the nucleus. As a result, the 3s electronexperiences a nuclear charge greater than +1e more oftenthan does a 3p electron, and (Zeff)3s (Zeff)3p.


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Figure 2

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

(8)

This equation shows that the dependence of E on arisesfrom the dependence of Zeff on .

Since (Zeff)3s > (Zeff)3p, the 3s orbital has a lower energythan the 3p orbital. This is clearly consistent with thearguments presented above; the 3s electron feels a largereffective nuclear charge, is therefore bound more tightly tothe nucleus, and thereby has a lower (more negative) energythan the less tightly bound 3p electron.

In the second part of this experiment, you will measure thespectrum of sodium and determine the wavelength of theemission line. From this wavelength, the effective nuclearcharges 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 yellowcast of low-energy sodium streetlights. This line arises fromthe transition of an electron from the excited electronic statein which the valence electron is in a 3p orbital to the groundelectronic state in which the valence electron is in a 3sorbital. By measuring the sodium spectrum, you will be ableto determine the energy difference between these twoelectronic states and thereby the energy difference betweenthe 3p and 3s orbitals. The existence of this emission lineshows that electrons in the 3s and 3p orbitals are of differentenergy and that their energy depends on the , as well as then, quantum number.

To determine the absolute energies of the sodium 3s and 3porbitals, additional information is required. This is providedby the ground state ionization energy, which is the energyrequired to remove the 3s valence electron from the groundelectronic state of the sodium atom. That is, the ionizationenergy is the energy of a transition from the 3s level to then= level. This is shown schematically in Figure 3.


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Figure 3Schematic of Sodium Atom Energy Levels

The ionization energy of gaseous sodium atoms is 496 kJmol-1, or 8.32 10-19 J for a single sodium atom. This valuecan be used with the wavelength of the sodium D line todetermine absolute energy values for the 3s and 3p levels.The wavelength of emitted light corresponds to the differencebetween the 3s and 3p orbital energies. The ground stateionization energy (IE) is the energy required to transfer thevalence 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 ED line = Efinal - Einitial = E3s - E3p (10)

therefore E3p = E3s - ED line (11)

Recall, that for an emission process, E is negative, since theatom loses energy. Thus, the above equation can be written inthe alternate form:

E3p = E3s + | ED line | (12)

Once the absolute energy of an orbital and its quantumnumber n are known, Zeff can be calculated using equation(8). For the sodium portion of this experiment, you willdetermine the wavelength of the D line, convert this into anenergy, ED line and calculate the energies and the Zeff valuesfor the 3s and 3p orbitals.


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The Meterstick Spectroscope and the DiffractionEquationIn this experiment, a simple spectroscope made frommetersticks will be used to observe atomic spectra. The lightis supplied by a gas discharge tube, which works like a neonsign. A sample of gas is sealed inside a glass envelope, withelectrodes in it. A high voltage is applied across theelectrodes and a plasma is formed, with free, acceleratedelectrons dissociating the hydrogen molecules into excitedatoms. These excited atoms emit light, as electrons in excitedelectronic states make transitions to electronic levels of lowerenergy. A diffraction grating is used to separate the emittedlight into its component wavelengths and a meterstick is usedto measure the positions of the emitted lines of light.

A schematic diagram of the meterstick spectroscope is shownin Figure 4. Light from the discharge tube passes through acollimating slit and the incident beam is transmitted through adiffraction grating. A transmission diffraction grating is madeby cutting equally spaced parallel grooves (also calledrulings) in a glass plate. The incident beam of light isdiffracted by the rulings on the grating and emission lines canbe viewed along the meterstick, on either side of the incidentbeam, as indicated by observers 1 and 2 in Figure 4. Emissionlines of different wavelengths are diffracted at differentangles, , 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 theslit was diffracted a different angle and each has a differentwavelength. The diffraction equation discussed in lecture andreproduced below can be used, with the distances from theslit to the observed line and from the slit to the grating todetermine the wavelengths of the observed lines of light.

Figure 4Schematic of the Meterstick Spectroscope, seen from

above.


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Arrows indicate the path of light.

A derivation of the fundamental diffraction equation requiredfor analysis of the spectral data was given in lecture and isreproduced here. In order for constructive interference tooccur at the angle , waves from the upper ruling on thediffraction grating must be in phase with waves from thelower ruling, as shown in Figure 5.

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

Figure 6 shows that this is possible if the path difference TScorresponds to an integral number of wavelengths, :

Figure 6Close-up of the diffraction of light by one ruling on thegrating. The path difference is TS.

That is,

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

A consideration of the right-angle triangle RST shows that

sin = TS/d (14)

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


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

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

In this experiment, you will observe the first-order diffractionpattern, so that m is always equal to 1. Thus, the diffractioncondition reduces to = d sin . From this equation it shouldbe clear that for a given value of d (i.e. for a given diffractiongrating), the angle at which constructive interferenceoccurs will depend on the wavelength, , of the emittedradiation. Conversely, this equation shows that themeasurement of the angle leads directly to a calculation ofthe wavelength, . This diffraction angle, , can bedetermined from the position of the diffracted emission lineson the meterstick, as shown in Figure 7. You may wish toconvince yourself of this geometry.

Figure 7Geometry of the Meterstick Spectroscope

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

= arctan (a/b) (16)

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

= d sin (17) = d sin (arctan a/b)


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

How Does a Fluorescent Light Work?A fluorescent light operates like a discharge tube, but hasbeen optimized to give diffuse, white light in order to beeasier on the eyes. The tube of a fluorescent light bulbcontains a low pressure of gas, which emits visible and UVlight when a voltage is applied across the tube's electrodes.The inside of the tube is coated with a phosphorescentmaterial that re-emits this light at wavelengths throughout thevisible region, making the light from the lamp appear white.In the final part of this experiment, you will COMPAREthe spectrum of a fluorescent light to that of severalelements, to determine what gas is inside the fluorescenttube.

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