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4000 4500 5000 5500 6000 6500 7000 7500. When a gas in a tube is subjected to a voltage, the gas ionizes, and emits light. Topic 7.1 Extended C – The Bohr theory of the hydrogen atom. We can analyze that light by looking at it through a spectroscope.
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4000 4500 5000 5500 6000 6500 7000 7500 When a gas in a tube is subjected to a voltage, the gas ionizes, and emits light. Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom We can analyze that light by looking at it through a spectroscope. A spectroscope acts similar to a prism, in that it separates the incident light into its constituent wavelengths. For example, barium gas in a gas discharge tube will produce an emission spectrum that looks like this: The emission spectrum is really an elemental fingerprint - it uniquely identifies the element producing it. FYI: The wavelengths are given in angstroms Å. 1 Å = 10-10 m.
4000 4500 5000 5500 6000 6500 7000 7500 4000 4500 5000 5500 6000 6500 7000 7500 FYI: Note the possible fingerprint for calcium as an absorbing constituent of the Sun's outer atmosphere. Calcium gas produces this spectrum: Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom Not only do glowing gases emit spectral lines, but cool gases absorb light in the same wavelengths and produce what is called an absorption spectrum. For example, the Sun produces a continuous spectrum that looks like this... ...with characteristic absorption spectral lines, revealing what non-glowing elements are in the Sun's atmosphere.
0 200 400 600 800 1000 1200 1400 1600 1800 2000 FYI: Balmer studied the middle series, which has parts in the visible spectrum. Obviously, the series is in his name. In the late 1800s a Swedish physicist by the name of J.J. Balmer observed the spectrum of hydrogen - the simplest of all the elements: Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom Emission Spectra of Hydrogen Paschen Series IR Balmer Series Lyman Series UV The general spectral signature is divided up into natural groups of spectral lines called series, falling roughly in the UV (ultraviolet), Visible, and IR (infrared) ranges of wavelengths. FYI: Each series is characterized by spectral line "bunching" at the smaller wavelengths, and "spreading" at the larger wavelengths.
Balmer Series Balmer Series In fact, Balmer found an empirical formula that predicted the allowed spectral wavelengths for the Balmer series of the hydrogen atom: Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom 1 n2 1 22 1 = R - for n = 3,4,5,... where R = 1.09710-2 nm-1 is called the Rydberg constant. FYI: The visible spectrum for hydrogen was found to fit this formula, but it was NOT understood why. FYI: Similar formulas were found to fit the other two series.
+ Energy in H Atom r FYI: Note that we are using the relation U = -kqQ / r for two point charges. Recall that this energy is negative since q and Q are oppositely charged. FYI: Up to this point in Bohr's derivation, classical mechanics has been used. An explanation was finally given in 1913 by the Danish physicist Niels Bohr. Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom Bohr postulated that the single electron was held in a circular orbit about the single proton in the hydrogen nucleus by the Coulomb force: Fc = FE ke2 2r ke2 r2 mv2 r 1 2 mv2= = The total mechanical E energy of the hydrogen atom is given by E = K + U so that ke2 r 1 2 E = mv2- ke2 2r 2 2 ke2 2r ke2 r E = - E = -
Principal Quantum Number - H Atom + r Allowed Radii - H Atom FYI: We subscript the r to indicate that the electron can only orbit the nucleus at certain quantized radii, determined by the principal quantum number. Recall that the angular momentuml of a point mass moving in a circle of radius r is given by Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom l = mvr Bohr then postulated the radical idea that the angular momentum of the electron was quantized, just like light. He stated that the angular momentum of the electron can only carry the discrete values given by h 2 mvr = n for n = 1,2,3,... ke2 2r 1 2 and the previous equation we can eliminate v, solving for r: From mv2= h2 42ke2m for n = 1,2,3,... r = n2 n
Energy Quantization in the Bohr Hydrogen Atom The Bohr Hydrogen Atom We can then take our energy equation for the hydrogen atom and substitute our allowed values for r: ke2 2r Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom E = - 22k2e4m h2 1 n2 for n = 1,2,3,... En = - Everything in the parentheses is a constant whose value we know. We can then rewrite both rn and Enlike this: -13.6 n2 En = eV rn = 0.0529n2 nm
FYI: We call the lowest energy level (n = 1) the GROUND STATE. We call the next highest energy level (n = 2) the 1ST EXCITED STATE. We call the next highest energy level (n = 3) the 2ND EXCITED STATE. Et cetera. What are the orbital radius and energy of an electron in a hydrogen atom characterized by principal quantum number 3? -13.6 n2 Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom En = eV rn = 0.0529n2 nm -13.6 32 r3 = 0.052932 nm E3 = eV r3 = 0.4761 nm E3 = -1.51 eV What is the change in energy if the electron "drops" to the energy characterized by principal quantum number 2? -13.6 22 E = (-3.4 - -1.51) eV E2 = eV E = -1.89 eV E2 = -3.4 eV FYI: In general, if an electron "drops" from a higher quantum state to a lower one, the hydrogen atom experiences a net loss of energy.
What is the orbital velocity of an electron in the second excited state (n = 3)? From the previous slide r3 = 0.4761 nm. Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom h 2 Then mvr = n nh 2mr v = 3(6.6310-34) 2(9.1110-31)(0.476110-9) v = v = 7.30105 m/s What then is the centripetal acceleration of the electron? (7.30105)2 0.476110-9 v2 r = 1.12 1021 ms-2 = ac = FYI: Classical theory predicts that electromagnetic radiation is created by accelerating charges. Since the hydrogen atom only radiates when its electron "drops" from one excited state to a less energetic state, Bohr postulated that "the hydrogen electron does NOT radiate energy when it is in one of its bound states (allowed by n). It only does so when "dropping" from a higher state to a lower state."
EXAMPLE: If a photon having an energy of E = +12.09 eVis absorbed by a hydrogen atom in its ground state, the electron will "jump" up to the second excited state (n = 3) sinceE = (-1.51 - -13.6) = +12.09 eV. n = 0.00 eV Consider a plot of energies for n = 1 to : n = 5 -0.544 eV n = 4 -0.850 eV Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom Excited States -13.6 n2 n = 3 -1.51 eV En = eV FYI: Bohr's theory only allows electrons in the hydrogen atom to absorb or emit photons having energies equal to the difference between any two of the allowed states shown here. n = 2 -3.40 eV Ground State n = 1 -13.6 eV EXAMPLE: If an electron at r3 suddenly "drops" to the ground state, the hydrogen atom LOSES energy having a value of E = (-13.6 - -1.51) = -12.09 eV. To conserve energy, a PHOTON having an energy of E = +12.09 eV is released.
0 200 400 600 800 1000 1200 1400 1600 1800 2000 FYI: The Lyman Series has as its final state the GROUND STATE. The Balmer Series has as its final state the FIRST EXCITED STATE. The Paschen Series has as its final state the SECOND EXCITED STATE. n = 0.00 eV n = 5 -0.544 eV So how do the three series of hydrogen spectra relate to the Bohr model? n = 4 -0.850 eV Second Excited State n = 3 -1.51 eV Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom Paschen Series IR Consider the following transi-tions of hydrogen from higher to lower states: First Excited State n = 2 -3.40 eV Balmer Series Visible Each transition gives off a photon of a different wavelength. Lyman Series UV Ground State n = 1 -13.6 eV Paschen Series IR Balmer Series Lyman Series UV