One line, element after element
In 1913 Henry Moseley, a young physicist at Manchester, measured the Kα line of one element after another. He made each element the target of an X-ray tube. He measured the wavelength with a crystal, by the method of the next lesson. Then he looked for a pattern in the frequencies \(\nu = c/\lambda\).
The frequency rises steadily from each element to the next. It does not grow in proportion to \(Z\), though. It grows faster, roughly as \(Z^2\), so a graph of \(\nu\) against \(Z\) is a curve. Moseley plotted the square root of the frequency instead. The points fell on a straight line.
Read the straight line of Fig 4.1. It does not pass through the origin. It meets the \(Z\) axis at \(Z = 1\), so \(\sqrt{\nu}\) is proportional to \(Z - 1\), not to \(Z\). The copper point, at \(Z = 29\), has \(\sqrt{\nu} = 1.39 \times 10^{9}\,\text{Hz}^{1/2}\). So the slope is \(a = 1.39 \times 10^{9}/28 = 4.96 \times 10^{7}\,\text{Hz}^{1/2}\).
For a given characteristic line, the square root of its frequency is a linear function of the atomic number of the element: \(\sqrt{\nu} = a(Z - b)\). The constants \(a\) and \(b\) depend on the line; for the Kα lines \(b \approx 1\).
For two elements and the same line, \(a\) cancels. With \(b = 1\), \(\nu \propto (Z - 1)^2\). Since \(\lambda = c/\nu\), the wavelength goes as \(1/(Z - 1)^2\). So one known line gives the same line of any other element by a simple ratio.
Copper (\(Z = 29\)) has its Kα line at \(1.54\,\text{Å}\). Use Moseley's law to find the Kα wavelength of molybdenum (\(Z = 42\)).
(a) \(\lambda \propto 1/(Z - 1)^2\), so \(\lambda_{\text{Mo}} = \lambda_{\text{Cu}} \times \left(\dfrac{29 - 1}{42 - 1}\right)^2 = 1.54 \times \left(\dfrac{28}{41}\right)^2\).
(b) \((28/41)^2 = 0.466\), so \(\lambda_{\text{Mo}} = 0.72\,\text{Å} = 0.072\,\text{nm}\).
(c) The measured value is \(0.071\,\text{nm}\). The simple law with \(b = 1\) is right to about 1 %.
Values used: \(b = 1\) for Kα, \(1\,\text{Å} = 0.1\,\text{nm}\).
The Kα line of an unknown element has \(\sqrt{\nu} = 1.64 \times 10^{9}\,\text{Hz}^{1/2}\). Take \(a = 4.96 \times 10^{7}\,\text{Hz}^{1/2}\) and \(b = 1\). Find the atomic number of the element.
Find the ratio of the Kα frequency of zinc (\(Z = 30\)) to that of calcium (\(Z = 20\)). Give it as a decimal.
Why \(Z - 1\)
Moseley saw at once why his law holds. Kα is a fall from the L shell (\(n = 2\)) into the K shell (\(n = 1\)). In Bohr's model such a fall gives \(13.6\,Z^2(1 - 1/4)\) electron volts, which grows as \(Z^2\). Moseley's law says \((Z - 1)^2\). So the falling electron behaves as if the nucleus had one proton fewer.
The reason is the second K electron. Before the fall, the K shell holds one electron and one vacancy. The remaining K electron lies much closer to the nucleus than the L electron does. Seen from the L electron, it cancels about one unit of the nuclear charge. This is screening. The falling electron feels an effective charge of about \((Z - 1)e\).
\[ E_n = -\dfrac{13.6\,Z_{\text{eff}}^2}{n^2}\ \text{eV} \]
Why. the Bohr energy of a level of a one-electron atom with nuclear charge \(Z_{\text{eff}}e\).
\[ h\nu = E_2 - E_1 = 13.6\,Z_{\text{eff}}^2\left(1 - \dfrac{1}{4}\right)\ \text{eV} \]
Why. the Kα photon is a fall from \(n = 2\) to \(n = 1\).
\[ h\nu = 10.2\,Z_{\text{eff}}^2\ \text{eV} \]
Why. \(13.6 \times 3/4 = 10.2\).
\[ h\nu = 10.2\,(Z-1)^2\ \text{eV} \]
Read it. the remaining K electron screens one unit of charge, so \(Z_{\text{eff}} = Z - 1\).
Divide by \(h\) to get the frequency. Write \(13.6\,\text{eV}/h = Rc\), where \(R\) is the Rydberg constant. Then \(\nu = \tfrac34 Rc\,(Z - 1)^2\), and taking the square root gives Moseley's law with \(b = 1\). The slope is \(a = \sqrt{3Rc/4}\). With \(Rc = 3.28 \times 10^{15}\,\text{Hz}\), \(a = 4.96 \times 10^{7}\,\text{Hz}^{1/2}\), exactly the slope of the measured line.
Use \(E = 10.2\,(Z - 1)^2\) eV to find the Kα energy and wavelength of copper (\(Z = 29\)) and of molybdenum (\(Z = 42\)). Compare with the measured \(0.154\) and \(0.071\,\text{nm}\).
Copper: \(Z - 1 = 28\). Molybdenum: \(Z - 1 = 41\).
\(E_{\text{Cu}} = 10.2 \times 28^2\) eV, \(E_{\text{Mo}} = 10.2 \times 41^2\) eV, and \(\lambda = 1240/E\) nm.
Copper: \(E = 7997\,\text{eV} \approx 8.0\,\text{keV}\), \(\lambda = 0.155\,\text{nm}\). Molybdenum: \(E = 17146\,\text{eV} \approx 17.1\,\text{keV}\), \(\lambda = 0.0723\,\text{nm}\).
Copper agrees with the measured \(0.154\,\text{nm}\) to under 1 %. Molybdenum's \(0.0723\,\text{nm}\) is about 2 % longer than the measured \(0.071\,\text{nm}\). Values used: \(hc = 1240\,\text{eV nm}\), ground state of hydrogen \(-13.6\,\text{eV}\).
Moseley's formula with \(b = 1\) is a very good approximation, not an exact law. For molybdenum it gives \(0.0723\,\text{nm}\), while the measured Kα line is at \(0.071\,\text{nm}\). This book uses the measured molybdenum values (Kα \(0.071\,\text{nm}\), Kβ \(0.063\,\text{nm}\), K edge \(20.0\,\text{keV}\)) whenever a spectrum is drawn. Use \(10.2\,(Z - 1)^2\) only when a problem asks for it.
The Kα line of an unknown target is at \(0.076\,\text{nm}\). Use \(E = 10.2\,(Z - 1)^2\) eV and \(hc = 1240\,\text{eV nm}\) to find its atomic number.
Why does the Kα form of Moseley's law contain \(Z - 1\) and not \(Z\)?
For the Lα lines the same idea works with a bigger screening. An electron falling into the L shell has the two K electrons and several L electrons partly between it and the nucleus. Moseley found \(b \approx 7.4\) for the Lα lines, with a smaller slope. Each line has its own \(a\) and \(b\), but every line gives a straight line.
A roll call of the elements
Before Moseley, the elements were placed in the periodic table roughly in order of atomic mass. A few pairs refused to fit. Cobalt (\(58.93\) u) is heavier than nickel (\(58.69\) u), yet its chemistry puts it first. Mendeleev had simply swapped such pairs, and nobody could say why. The number that fixed the order was not yet known.
Moseley's X-rays settled it. Each step along his straight line is one unit of \(Z\), one more proton in the nucleus. Cobalt's Kα line is at \(6.93\,\text{keV}\) and nickel's at \(7.48\,\text{keV}\). So cobalt has \(Z = 27\) and nickel \(Z = 28\), in the order chemistry wanted. The atomic number, not the mass, orders the table.
The line also showed where elements were missing. Moseley's points jumped a step at \(Z = 43\), \(61\) and \(75\), among others. Each gap was a place with no known element. Rhenium (\(75\)) was found in 1925, technetium (\(43\)) was made in a laboratory in 1937, and promethium (\(61\)) was found in 1945.
Cobalt has a slightly larger atomic mass than nickel, but a Kα line of lower frequency. What does this show?
Many jewellers and hallmarking centres in India test gold with an X-ray fluorescence analyser. The machine shines X-rays on a ring or a chain. The atoms of the metal are left with inner-shell vacancies, and each element answers with its own characteristic lines. The analyser reads the lines of gold, silver, copper and zinc and reports the share of each. A 22-carat chain should show about 91.6 % gold.
Henry Moseley was 25 when he measured the X-ray lines of the elements in Rutherford's laboratory at Manchester in 1913. His straight lines gave every element its number and found the gaps in the table within a few months. When the First World War began, he joined the army as a signals officer. He was killed by a sniper at Gallipoli on 10 August 1915, at the age of 27. Many scientists felt that he would have won a Nobel Prize.
- Using \(\nu \propto Z^2\) for the Kα lines.The Kα law has \(Z - 1\): \(\nu \propto (Z - 1)^2\). Subtract one before squaring.
- Taking \(\lambda \propto (Z - 1)^2\).The frequency grows as \((Z - 1)^2\), so the wavelength falls as \(1/(Z - 1)^2\). Heavier targets give shorter lines.
- Using \(b = 1\) for every line.\(b \approx 1\) is for Kα. Other lines, such as Lα, have their own, larger \(b\).
- Treating \(10.2\,(Z - 1)^2\) eV as exact.It is good to about 2 %. Use measured line values when a problem gives them.