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CHAPTER 4 Structure of the

 4.1 The Atomic Models of Thomson and Rutherford  4.2 Rutherford  4.3 The Classic Atomic Model  4.4 The of the Atom  4.5 Successes and Failures of the Bohr Model  4.6 Characteristic X-Ray Spectra and  4.7 Atomic Excitation by 4.1 The Atomic Models of Thomson and Rutherford

Pieces of evidence that scientists had in 1900 to indicate that the atom was not a fundamental unit:

1) There seemed to be too many kinds of , each belonging to a distinct chemical element. 2) Atoms and electromagnetic phenomena were intimately related. 3) The problem of (원자가). Certain elements combine with some elements but not with others, a characteristic that hinted at an internal atomic structure. 4) The discoveries of radioactivity, of x rays, and of the Thomson’s Atomic Model

 Thomson’s “plum-pudding” model of the atom had the positive charges spread uniformly throughout a sphere the size of the atom with, the newly discovered “negative” electrons embedded in the uniform background.

 In Thomson’s view, when the atom was heated, the electrons could vibrate about their equilibrium positions, thus producing electromagnetic radiation. Experiments of Geiger and Marsden

 Under the supervision of Rutherford, Geiger and Marsden conceived a new technique for investigating the structure of matter by scattering particles (He nuclei, q = +2e) from atoms.

 Plum-pudding model would predict only small deflections. (Ex. 4-1)

 Geiger showed that many  particles were scattered from thin -leaf targets at backward angles greater than 90°.

Rutherford: ”It was as if you fired a 15 inch shell at a tissue paper and it came back and hit you” Example 4.1

 The maximum scattering angle corresponding to the maximum momentum change  Maximum momentum change of the α particle is or

 Determine θ by letting ∆pmax be perpendicular to the direction of motion. Multiple Scattering from Electrons?

 If an α particle were scattered by many electrons and N electrons results in  The number of atoms across the thin gold layer of 6 × 10−7 m:

 Assume the distance between atoms is

and there are

That gives Rutherford’s Atomic Model

 even if the α particle scattered from all 79 electrons in each atom of gold  2300 79(0.016oo )  6.8 total

The experimental results were not consistent with Thomson’s atomic model.

 Rutherford proposed that an atom has a positively charged core (nucleus) surrounded by the negative electrons. 4.2

The Assumptions 1. The scatterer is so massive that it does not recoil significantly; therefore the initial and final kinetic energies of the  particle are practically equal. 2. The target is so thin that only a single scattering occurs. 3. The bombarding particle and target scatterer are so small that they may be treated as point and charges. 4. Only the Coulomb force is effective. The Relationship Between the Impact Parameter b and the Scattering Angle 

 There is a relationship between the impact parameter b and the scattering angle θ.

When b is small,  r gets small.  Coulomb force gets large.  θ can be large and the particle can be repelled backward.

(p. 133-134) • however, cannot pick impact parameter in an experiment.

• particles are incident at varied impact parameters all around the scatterer.

• Any particle inside a circle of area 2 πb0 will be scattered at angles greater than θ0.

2 • Define σ = πb0 for scattering at an angle greater than θ0.

• σ related to the probability for a 2 particle being scattered by the σ = πb0 nucleus. Rutherford Scattering equation

The number of particles scattered per unit area is The Important Points

1. The scattering is proportional to the square of the

atomic number of both the incident particle (Z1) and the target scatterer (Z2). 2. The number of scattered particles is inversely proportional to the square of the of the incident particle. 3. For the scattering angle  , the scattering is proportional to 4th power of sin( /2). 4. The Scattering is proportional to the target thickness for thin targets. 4.3: The Classical Atomic Model As suggested by the the atom consisted of a small, massive, positively charged nucleus surrounded by moving electrons. This then suggested consideration of a planetary model of the atom. Let’s consider atoms as a planetary model. The Planetary Model is Doomed

 From classical E&M theory, an accelerated radiates energy (electromagnetic radiation) which means total energy must decrease. Radius r must decrease!!

Electron crashes into the nucleus!?  Obviously most atoms are stable, so once again classical breaks down at small length scales.  Physics had reached a turning point in 1900 with Planck’s hypothesis of the quantum behavior of radiation. 4.4: The Bohr Model of the

• In 1913 Neil’s Bohr developed a structural model for the hydrogen atom consistent with the nuclear model of Rutherford and the observed spectral series.

• Believed quantum principles should govern more phenomena than just blackbody radiation and the photoelectric effect.

• Bohr assumed electron moved around a massive positively charged nucleus ( of nucleus taken to be infinite).

• radius of the orbit >> radius of nucleus. The Bohr Model of the Hydrogen Atom

• Bohr also made the following general assumptions:

1) “Stationary states” (orbiting electrons in atoms do not radiate energy). These states have definite total energy. 2) EM radiation emitted/absorbed when electrons make transitions

between stationary states such that hf = E = E1 − E2 3) Classical laws of physics do not apply to transitions between stationary states. 4) The mean kinetic energy of the electron-nucleus system is quantized

such that K = n h forb/2, where forb is the frequency of rotation. As we shall see this is equivalent to quantizing the angular momentum of the stationary states in multiples of h/2π.

• Used these four assumptions to derive the Rydberg equation for characteristic spectra.

• Four assumptions keep as much as possible the ideas of classical physics. Quantization of angular momentum Bohr Radius

 Rearranging for radius, result is that the radii of the stationary states in the hydrogen atom are quantized:

Where the Bohr radius is given by

 The smallest diameter of the hydrogen atom is 

 plugging the radius into our previous expression in the planetary model for the energy of a stable electron orbit:

 n = 1 gives its lowest energy state (called the “ground” state) The Hydrogen Atom  The energies of the stationary states

where E0 = 13.6 eV

 Emission of light occurs when the atom is in an excited state and decays to a lower

energy state (nu → nℓ).

where f is the frequency of a photon.

R∞ is the Rydberg constant. Transitions in the Hydrogen Atom

Lyman series: n = 1 (invisible). Balmer series: n = 2 (visible). Paschen series: n = 3 (visible).

It was clear Bohr’s model predicts all the frequencies in atomic hydrogen. Fine Structure Constant

 The electron’s velocity in the Bohr model:

 On the ground state, 6 v1 = 2.2 × 10 m/s ~ less than 1% of the speed of light

 The ratio of v1 to c is the fine structure constant.

 α is a fundamental physical constant characterizing the strength of the electromagnetic interaction The Correspondence Principle

Classical electrodynamics + Bohr’s atomic model

Determine the properties of radiation

Need a principle to relate the new modern results with classical ones.

In the limits where classical and Bohr’s correspondence principle quantum theories should agree, the quantum theory must reduce the classical result. The Correspondence Principle

 The frequency of the radiation emitted fclassical is equal to the orbital frequency forb of the electron around the nucleus.

 The frequency of the transition from n + 1 to n is

 For large n,

Substitute E0: 4.5: Successes and Failures of the Bohr Model

 The electron and hydrogen nucleus actually revolved about their mutual center of mass.

 The electron mass is replaced by its reduced mass.

 The Rydberg constant for infinite nuclear mass is replaced by R.

111E0 RR22,   nnlu hc

 Discovery of Deuterium (1932) from slight splitting of Hα line between Hydrogen (656.47 nm) and Deuterium (656.29 nm) Limitations of the Bohr Model

The Bohr model was a great step of the new quantum theory, but it had its limitations.

1) Works only to single-electron atoms 2) Could not account for the intensities or the fine structure of the spectral lines 3) Could not explain the binding of atoms into molecules

➡ To overcome these limitations a full quantum mechanical treatment is required (Chapter 7). 4.6: Characteristic X-Ray Spectra and Atomic Number

 Discussed the production of X-rays previously (Section 3.7).  The bremsstrahlung radiation is superimposed on peaks whose wavelength is element specific.  Although strictly speaking the Bohr model only applies to one electron atoms it was believed that the Bohr-Rutherford atom would also apply to many-electron atoms.  Bohr model suggests that the

electrons occupy shells of radius rn characterized by the principle quantum number n. Characteristic X-Ray Spectra and Atomic Number Atomic Number

L shell to K shell Kα x ray

M shell to K shell Kβ x ray

 Atomic number Z = number of protons in the nucleus  Moseley found a relationship between the frequencies of the characteristic x ray and Z.

This holds for the Kα x ray Moseley’s Empirical Results

 The x ray is produced from n = 2 to n = 1 transition.  From Bohr model:

2  E0 and thus R is prop to (Ze) .

 For electron making a transition into K-shell from L-shell it feels an effective charge Z-1 due to remaining electron in the K-shell therefore:

 Moseley’s research clarified the importance of the electron shells for all the elements, not just for hydrogen.  Also demonstrated Z determined ordering in the . 4.7: Atomic Excitation by Electrons

 Franck and Hertz studied the phenomenon of ionization via electron bombardment.

 Observations:

 Accelerating voltage is below 5 V  electrons did not lose energy

 Accelerating voltage is above 5 V  sudden drop in the current Atomic Excitation by Electrons

 Ground state has E0 to be zero.

First excited state has E1.

The energy difference E1 − 0 = E1 is the excitation energy.

 Hg has an excitation energy of 4.88 eV in the first excited state

 No energy can be transferred to Hg below 4.88 eV because not enough energy is available to excite an electron to the next energy level

 Above 4.88 eV, the current drops because scattered electrons no longer reach the collector until the accelerating voltage reaches 9.8 eV and so on. Atomic Excitation by Electrons

 The Franck-Hertz experiment provided further proof of the quantization of atomic energy levels.

 Bombarding electron kinetic energy can only change in discrete amounts which depend on the energy levels of the atoms in the vacuum tube.