chemlab - chemistry 6 - spectrum of the hydrogen atom - chemistry & background

8/12/2019 ChemLab - Chemistry 6 - Spectrum of the Hydrogen Atom - Chemistry & Background

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Spectrum of theHydrogen Atom

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The line spectra of the elements, like those observed in this

experiment, show that electrons in atoms can onlyexist with

discrete, quantizedenergy 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 energyare called excited electronic states. An electron in an excited

electronic state can loseenergy 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 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 also limited to particular energy values. Thus,

transitions between electronic energy levels, observed either asemission 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 differencesbetween energy levels of the electron in

a hydrogen atom.

The hydrogen emission spectrum consists of several series of

lines, named for their discoverers. A seriesof emission lines

consists of those electronic transitions which allterminate 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
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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, meis the

mass of the electron, ois 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 nucleusthan 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


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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 emittedcorresponds to the energy level difference between 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, this equation

expresses the differencebetween 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 emissionprocess, the atom losesenergy. Its

energy becomes morenegative and E for the atom is negative.

This is consistent with the above equation since in an emission

process nfinalis less than ninitial. In an absorption process, the

atom gainsenergy, nfinalis greater than ninitialand E 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 10-18J) ( 1/22 - 1/32) (6)

= (2.178 10-18J) ( 1/4 - 1/9)

= 3.035 10-19J

As mentioned above, since this is an emission process, E for the


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atom is negative. Thus, the energy gained by the surroundings, i.e.

the energy of the emitted photon, is given by |E |, and this can

be converted to the wavelength of light emitted using the

relationship between energy and the wavelength of light, E = h c/

:

= h c / | E | (7)

= 6.5452 10-7m = 654.52 nm

Thus, the lowest energyline (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 ChargeThe

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

Schrdinger confirms the correctness of the Bohr energy level

expression for one-electron atoms: for one-electron atoms the

energy of the electron depends onlyon the value of the principal

quantum number n:

For multielectronatoms, quantum mechanics shows that energy

levels in such systems are quantized and that the energy of an

electronic level depends on bothn 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 samevalue 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 alwayscloser to the

nucleus than the 3s valence electron, the 3s electron would

alwaysexperience a nuclear charge of +1e, which is the +11e of

the nucleus combined with the 10e charge of the other electrons.


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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 greaterthan +1e. At such instants, the 3s

electron is said to be imperfectly shielded from the full nuclearcharge. 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 plotsin 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 greaterthan

+1e more often than does a 3p electron, and (Zeff)3s(Zeff)3p.


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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 Zeffon .

Since (Zeff)3s> (Zeff)3p, the 3s orbital has a lower energy than


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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 emissionline. 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 toremove 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.


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Figure 3

Schematic of Sodium Atom Energy Levels

The ionization energy of gaseous sodium atoms is 496 kJ mol-1,

or 8.32 10-19J 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 3senergy level to the n= level, which is defined as the zero of

energy. Thus,

IE =Efinal- Einitial=En=- E3s=0 - E3s= -E3s (9)

andED line= Efinal- Einitial= E3s- E3p (10)

therefore E3p= E3s- ED line (11)

Recall, that for an emission process, E is negative, since the

atom loses energy. Thus, the above equation can be written in the

alternate form:

E3p= E3s+ | ED line| (12)

Once the absolute energy of an orbital and its quantum number n


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are known, Zeffcan 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, ED lineand

calculate the energies and the Zeffvalues 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 meterstickis 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 byobservers 1 and 2 in Figure 4. Emission lines of different

wavelengths are diffracted at different angles, , 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

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.


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Figure 4

Schematic of the Meterstick Spectroscope, seen from above.

Arrows indicate the path of light.

A derivation of the fundamental diffraction equation required foranalysis of the spectral data was given in lecture and is

reproduced here. In order for constructive interference to occur

at the angle , 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, :


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Figure 6

Close-up of the diffraction of light by one ruling on the grating.

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 on the

diffraction grating. Thus for constructive interference

m = d sin 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 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 = d sin . From this equation it should be

clear that for a given value of d (i.e. for a given diffraction

grating), the angle at which constructive interference occurs will

depend on the wavelength, , of the emitted radiation.

Conversely, this equation shows that the measurement of the

angle leads directly to a calculation of the wavelength, . This

diffraction angle, , 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.


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Figure 7

Geometry of the Meterstick Spectroscope

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

meterstick,

= arctan (a/b) (16)

To determine the wavelength, , of the observed line, this is

combined with the first-order diffraction condition to give

= d sin (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 pressureof 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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