Page 235 - Essentials of physical chemistry

Basic Spectroscopy 197

inner orbits. We have shown above that certain metal targets (Cu) can be selected to generate a
preponderance of K a x-rays for use in medical imaging although the Bohr model is too simple and
results in an error of about 6.2%. That implies that various elements have characteristic K a and L a
wavelengths that can be used for elemental analysis within our understanding that the model does
not include the effect of the repulsion between electrons. Our schematic in Figure 9.5 is oversim-
plified but shows some features that should be part of the science education of undergraduates.
The diagram shows how a beam of electrons can be accelerated to high velocity to form what is
called an ‘‘electron microscope’’ to form an image on film or on a fluorescent screen by transmission
of the electron beam accelerated through a potential of 75,000 V or even higher. An electron beam
can be focused using electrostatic lenses (not shown on the schematic) and the development of
electron microscopes has reached a high level of sophistication to visualize bacteria and viruses in
biological samples.
The abbreviation SEM indicates a ‘‘scanning electron microscope’’ that can use an x–y scan to
form a pixel image due to changes in transmitted intensity of the beam. However, in the last few
years it has been realized that the same device can be used with a lower voltage of the electron beam
of typically 20,000 V. The lower voltage is still sufficient to cause K a x-rays for elements up to
Z ¼ 48 (Cd) but the same LiF diffraction monochrometer and detector can then be used to measure
L a and L b XRF (x-ray fluorescence) since for elements with Z > 48, the L shell is much deeper in
energy. We can use the Bohr formula to estimate the L a wavelength also:

1 1 5
2
2
¼ E 2 E 3 ¼ (Z ) (13:6057 eV) ¼ Z 2 (13:6057 eV) ¼ Z (1:88968)eV
DE L a 2 2
2 3 36
We can use the experimental wavelengths of the L a transitions to probe the interior of the heavier
elements. We saw above that we could use the integer Z value of the nucleus for the K a wavelengths
of the (n ¼ 2) ! (n ¼ 1) transition of the Bohr atom and obtain reasonable results.
However, for the case of the effective nuclear charge Z eff for the excited outer electron in the
sodium flame test, we found a noninteger value due to incomplete screening of the nuclear charge by
the shell of electrons between the (n ¼ 3, n ¼ 4) shells of the atom and what we can call the
Neon-core of the inner part of the Na atom. In the case of the L a energy, the transition is from
the (n ¼ 3) ! (n ¼ 2) level, so the K shell is between the L shell and the bare nucleus for sure.
Further, the L shell may still be mostly occupied with perhaps one vacancy so the Ne core may be
mostly intact.
For elements heavier than Cd, we expect that the Bohr radius of the (n ¼ 1) is quite small so that
the screening (covering up) of the nuclear charge by the K shell should be close to 2 for the two
electrons in the K shell. However, the next eight electrons in the L shell may still be mostly there
with perhaps only one or two vacancies. Thus, the effective nuclear charge for the heavier elements
should be close to Z eff ¼ Z 8. Consider In where Z ¼ 49. From Table 9.1, we see that the
wavelength of the L a transition is 0.3772 nm ¼ 3.772 Å. From the formula above for the L a
transition energy, we can solve for the value of Z eff for the charge that is seen by the electron
falling into the L shell as

12,398 eV A ˚
2
¼ Z (1:88968)eV:
3:772 A ˚
DE L a ¼ E 2 E 3 ¼ eff
Thus, for In (Z ¼ 49), we find

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
12,398 eV A ˚
¼ 41:7057:
Z eff ffi
(3:772 A ˚ )(1:88968 eV)