Chapter 28
The Giancoli onLine Tutor

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Quantum Mechanics of Atoms

I. Angular Momentum in the Atom

In quantum Mechanics the physical description and understanding of the hydrogen atom is much different than that given by the Bohr picture but problem working is much the same. The energy levels are the same (E= -13.6(eV)/n²), one difference is that in the Bohr atom the angular momentum, L, of the electron is always equal to nh/2p. In quantum mechanics L = SQRT(l(l+1)), where l = 0, 1,...,n-1 and for each l, the allowed values of L_z are given by L_z = m_lh/2p where m_l = -l, -l+1,...,l.. This effectively allows the atom to have certain orientations in space as shown at the left.

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Electrons in Complex Atoms

More complicated atoms are thought of being made up of electrons in the various hydrogen states, filling from the bottom up. The electrons must each go in a separate state, that is, have different quantum numbers n, l, m_l and a new quantum number the spin quantum number, m_s, which can have only the values +1/2 and -1/2 ( 'spin-up' and 'spin-down'). This is called the Pauli Exclusion Principle

The electrons fill from the bottom . Electrons with the same value of n are said to be in the same shell (n=1->K, n=2->L, n=3->M,...) and those with the same value of n and l are in the same subshell labelled according to the value of l (l=0->s, l=1->p, l=2->d,...). As the atoms get larger the order of filling becomes less obvious, for example, the 4s subshell fills before the 3d. In the diagram at the right the configuration for Beryllium (Be) is shown with two in the 1s subshell and two more in the 2s subshell. The configuration would be given as 1s²2s².

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Interactive Activity

4s:

3d:

3p:

3s:

2p:

2s:

1s:

Click this button to obtain the number of electrons:
Number of electrons:
Enter what you think is the matching configuration on the right by entering x's in the appropriate boxes and then check your answer by clicking below.

4s:

3d:

3p:

3s:

2p:

2s:

1s:

II. The Uncertainty Principle

The Uncertainty Principle is the basis of many problems. It comes in two forms:
DxDp >=h/2p
and DEDt >=h/2p The first tells us that the uncertainty in the determination of the position of a particle multipied by the uncertainty in the determination of the momentum is greater than h/2p. The second tells us that the uncertainty in the determination of the energy of a particle multipied by the uncertainty in the time to determine it is greater than h/2p. h/2p has the value 1.055x10^-34 Js or 6.59x10^-16 eVs. The latter form is useful in working with the second equation.

A variation on the first form (for nonrelativistic particles) is given by: DxDv >=h/2mp
where v is the speed of the particle and m its mass.

Knowing how a particle is confined we can easily use the uncertainty principle to get a handle on its momentum or speed. This is based on the assumpion that the momentum ( or speed ) can not be less than the uncertainty in its value.

For example, an electron confined to a region the size of the Hydrogen atom (1x10^-10) will have an uncertainty in its momentum of, at least, Dp = 1.055x10^-34/1x10^-10 ~ 1x10^-24 kgm/s. If that was its momentum it would correspond to an energy of p²/2m =( 1x10^-24)²/(2*9.1x10^-31) = 5.5x10^-19 J. This corresponds to an energy of 3.4 electron-volts roughly the magnitude of the energy of an electron in a a hydrogen atom.

Excited atomic states such as the excited states (n=2, 3,...) of Hydrogen have an experimental spread in their energy because they exist only for a very small finite time. If the lifetime of an atomic state is 1x10^-9 s, what is the uncertainty in the energy in eV? DE = 6.59x10^-16/1x10^-9 ~ 6.6x10^-7 eV .

Interactive Examples
What is the uncertainty in the speed of an electron (m= 9.1x10^-31) to the nearest m/s, if it is confined to a distance of 1 micron (1x10^-6 m)? (Enter your answer and click outside)

A certain nuclear state as a uncertainty in energy of 100 eV, what is the lifetime of the state to the nearest attosecond (1x10^-18 second)? (Enter your answer and click outside)

III. X-rays

At the left an X-ray tube is illustrated. Electrons are accelerated through a potential difference of V and acquiring and energy of eV in the process before entering a metal target. Most undergo glancing collisions with the atoms in the target, when this happens they are accelerated and emit radiation (pictured here as red X-ray photons). Occasionally, an electron will undergo a 'head-on' collision and loose all its energy in one photon (here blue). This will be the highest energy, highest frequency and shortest wavelength (called cutoff wavelength) X-ray produced.

For this

E_photon = eV
hf_photon = eV
f_photon = eV/h
l_o = c/f_photon
l_o = hc/eV

Another happening occurs when an electon causes one of the inner (K shell) electrons in the atom to be ejected, the atom is now in an excited state (red in the above) and an upper electron falls into the vacated state emiting what is called a characteristic (characteristic of the atoms in the metal) X-ray photon (green above). Two are observed, and appear as peaks on the X-ray spectum (see above right). The K_a X-ray results from an electron in the L shell falling into the K shell vacancy, while the K_b X-ray is produced by an M shell electron making the transistion.

For a metal target with atomic number Z an electron in the n=2 shell (L shell) sees a positive charge of Z shielded by one negative charge (the remaining K shell electron). Applying a Bohr model to this situation, its energy will be
E₂ = -13.6(Z-1)²/2²
assuming that its final state can be represented by
E₁ = -13.6(Z-1)²/1²
The energy of the resulting photon will be the difference or
E_photon = 13.6(Z-1)²*3/4 = 10.2(eV)(Z-1)² = 1.63x10^-18(J)(Z-1)².

Thus if the target was iron (Z=26), the photon energy would be
E_photon = 10.2*25² = 6375 eV = 1x10^-15 J
The wavelength would be

l = c/f = hc/E = 6.63x10^-34*3x10⁸/1x10^-15 = .2 nm

Interactive Examples
An X-ray tube has an accelerating voltage of 50 kV. What is the wavelength of the most energetic photon to the nearest pm (1x10^-12) m? (Enter your answer and click outside)

If the target is molybdenum (Z= 42), what is the energy of the Ka X-ray to the nearest keV? (Enter your answer and click outside)

What is the wavelength to the nearest pm? (Enter your answer and click outside)