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The Bohr Model of Hydrogenic Ions

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<strong>The</strong> <strong>Bohr</strong> Atom 4<br />

This value agrees very well with the observed wavelength <strong>of</strong> the Lyman- α line, 1215.7 A, °<br />

indicating that the <strong>Bohr</strong> model may have some validity. But there still remains a thorny<br />

problem. If the electron radiates energy as it moves in its orbit, the electron must loose energy,<br />

and therefore spiral inward toward the nucleus, eventually collapsing. Thus, this picture is not<br />

complete.<br />

In an attempt to circumvent this classical collapse, <strong>Bohr</strong> looked to the findings <strong>of</strong><br />

Einstein and Planck. Since the energy <strong>of</strong> electromagnetic waves seem to come in packets <strong>of</strong><br />

size 2/ , <strong>Bohr</strong> postulated that the energy emitted from an atom must be emitted in dicrete<br />

chunks <strong>of</strong> energy equal to 2/ . Thus, <strong>Bohr</strong> postulated that an electron could exist is some<br />

orbiting state within the atom without radiating, but if the electron were to change from one<br />

stable state to another, the energy change in the atom must be equal to the energy emitted by<br />

the photon, or<br />

I I œ 2<br />

0 3 / (1.11)<br />

To continue this line <strong>of</strong> reasoning, we must look for an expression for the total energy <strong>of</strong><br />

the atom. This is accomplished by combining the kinetic energy <strong>of</strong> the electron and the<br />

electrostatic potential energy <strong>of</strong> the atom. <strong>The</strong> kinetic energy <strong>of</strong> the electron is related to the<br />

electrostatic force by Equation 1.6, from which we can write<br />

<strong>The</strong> electrostatic potential energy <strong>of</strong> the electron is given by<br />

which gives<br />

" 5Ð^/Ñ/<br />

7@ # œ<br />

(1.12)<br />

# # <<br />

Y Ð

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