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 = hf
Therefore:
X ray frequency (f) = eV/h
This shows that the maximum frequency is directly proportional to the accelerating voltage.
Example
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  (= 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 2.48 x 1015.
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- Square root of ½
frequency (√f)/Hz
shell) in the atom from other excited x10-8
levels give out X-rays in a series of
wavelengths like an optical spectrum.
30
This is known as the K-series, and individual lines are denoted by K, K and so
on. Electron transitions ending on the
20
second level are known as the L-series.
10
0
10
20
30
40
50
60
Atomic number (Z)
1
The following table shows the wavelengths of the K, lines for some elements:
Element
Aluminium
Calcium
Manganese
Iron
Cobalt
Nickel
Proton number
13
20
25
26
27
28
Wavelength (nm)
0.823
0.335
0.210
0.194
0.179
0.166
Element
Copper
Bromine
Silver
Tungsten
Uranium
Proton number
29
35
47
74
92
Wavelength (nm)
0.139
0.104
0.056
0.021
0.017
2