Calculation of X-ray wavelengths
If electrons are accelerated to a velocity v
by a potential difference V and then allowed to collide with a metal target, the maximum
frequency of the X-rays emitted is given by the equation:
½ mv2= eV =
hfTherefore:
X ray frequency (f) = eV/h
This shows that the maximum frequency is directly proportional to the accelerating voltage.
Example problem
Calculate the minimum wavelength of X-rays emitted when electrons accelerated through 30 kV strike a target.
f = [1.6 x 10-19 x 3 x 104]/ 6.63 x 10-34 = 7.2x1018 Hz
Therefore the wavelength l (= c/f) is 0.41 x 10-10 m = 0.04 1 nm (compared with some 600 nm for yellow light).
Moseley's law
In 1914 Moseley proposed a law showing how the X-ray frequency can be related to the proton (atomic) number Z of the
target material. If f is the X-ray frequency, then:
X ray frequency (f) = k(Z- b)2
where k and b are constants, k having a value of 2.48x10
15.
Plotting a graph of Z against √f will give a straight line as shown
in the diagram, and in fact Moseley predicted the existence of elements 43, 61, 72 and 75 by
the gaps that he found in his original version of the graph.
Electrons falling to the lowest level (or K-shell) in the atom from other
excited levels give out X-rays in a series of wavelengths like an optical spectrum. This is
known as the K-series, and individual lines are denoted by Ka, Kb and so on.
Electron transitions ending on the second level are known as the L-series.
The following table shows the wavelengths of the K
a lines for
some elements:
| Element |
Proton number |
Wavelength (nm) |
Element |
Proton number |
Wavelength (nm) |
| Aluminium |
13 |
0.823 |
Copper |
29 |
0.139 |
| Calcium |
20 |
0.335 |
Bromine |
35 |
0.104 |
| Manganese |
25 |
0.210 |
Silver |
47 |
0.056 |
| Iron |
26 |
0.194 |
Tungsten |
74 |
0.021 |
| Cobalt |
27 |
0.179 |
Uranium |
92 |
0.017 |
| Nickel |
28 |
0.166 |
|
|
|
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB
