Physical Principles of Electron Microscopy: An Introduction to TEM ...

156 Chapter 6
where h = 6.63 � 10 -34 Js is the Planck constant. Because the left-hand side
of Eq. (6.2) represents the angular momentum of the electron and because n
is any integer, known as the (principal) quantum number, Eq. (6.2)
represents the quantization of angular momentum. Without this condition,
the atom is unstable: the centripetal acceleration of the electron would cause
it to emit electromagnetic radiation and quickly spiral into the nucleus. This
problem does not arise in connection with planets in the solar system
because
they are electrically neutral.
Using Eq. (6.2) to substitute for v in Eq. (6.1) provides the radius rn of
the
orbit of quantum number n :
rn = n 2 (h/2�) 2 /(KmZe 2 ) = n 2 a0 /Z (6.3)
Here, a0 = 0.053 nm is the (first) Bohr radius, corresponding to the radius of
a hydrogen atom (Z = 1) in its lowest-energy ground state (n = 1). We can
now use Eq. (6.2) to solve for the orbital speed vn or, more usefully, employ
Eq. (6.1) to calculate the total energy En of an orbiting electron as the sum of
its
kinetic and potential energies:
En = mv 2 /2 � K(Ze)(e)/rn = KZe 2 /(2rn) � KZe 2 /rn
= � K Z e 2 /(2rn) = � R (Z 2 /n 2 ) (6.4)
Here, R = Ke 2 /(2a0) = 13.6 eV is the Rydberg energy, the energy needed to
remove the electron from a hydrogen atom (Z 2 = 1) in its ground state (n = 1)
and therefore equal to the ionization energy of hydrogen.
n=2
n=3
(a) (b)
M
L
E
0
K n = 1
n=3
n = 2
Figure 6-1. Bohr model of a carbon atom, visualized in terms of (a) electron orbits and (b) the
equivalent energy levels. The orbits are also described as electron shells and are designated as
K (n = 1), L (n = 2), M (n = 3) and so forth. The dashed arrows illustrate a possible scenario
for x-ray emission: a K-shell electron (nl = 1) is excited to the empty M-shell (nu = 3) and an
L-shell electron fills the K-shell vacancy in a de-excitation process (nl = 1, nu =2).
r