The Workbook

1. The electrons are stopped suddenly by the target atoms. A charge that decelerates radiates, and the radiation here is X-rays. Some electrons also knock inner electrons out of target atoms, which then emit characteristic X-rays.

2. A heavy nucleus decelerates the electrons more strongly, so more of their energy becomes X-rays. A heavy metal also has a high melting point.

3. About \(99\%\) of the electrons' energy becomes heat in the target. Without cooling the target would melt.

\(\lambda_{\min} = \dfrac{hc}{eV} = \dfrac{12400}{V}\) Å with \(V\) in volt.

1. \(\dfrac{12400}{12400} = 1\) Å.

2. \(\dfrac{12400}{31000} = 0.4\) Å.

1. The \(25\) kV curve starts at \(0.5\) Å and the \(50\) kV curve at \(0.25\) Å. Check: \(12400/25000 = 0.5\) Å and \(12400/50000 = 0.25\) Å.

2. The peaks are the characteristic lines of molybdenum. Their wavelengths depend on the energy levels of the target atom, not on the tube voltage.

3. The smooth hump is the continuous spectrum (bremsstrahlung). The two sharp peaks are the characteristic spectrum.

1. \(K = eV = 25\) keV.

2. The most energetic photon is made when an electron loses all its energy in one collision: \(25\) keV, which is \(\lambda_{\min} = 0.5\) Å.

3. Most electrons lose their energy in several steps, and each step gives a photon of only part of the energy. All values from zero to \(25\) keV occur, which makes the spectrum continuous.

1. \(\lambda_{\min} \propto \dfrac{1}{V}\), so it halves. The electrons are more energetic, so the intensity also rises.

2. More electrons per second hit the target, so the intensity rises. Their energy is unchanged, so \(\lambda_{\min}\) stays the same.

1. \(K = eV = 1.6 \times 10^{-19} \times 10^{4} = 1.6 \times 10^{-15}\) J.

   \(v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 1.6 \times 10^{-15}}{9.1 \times 10^{-31}}} = \sqrt{3.5 \times 10^{15}} \approx 5.9 \times 10^{7}\) m/s, about \(0.2c\).

2. \(\lambda_{\min} = \dfrac{12400}{10000} = 1.24\) Å.

1. \(n = \dfrac{I}{e} = \dfrac{0.02}{1.6 \times 10^{-19}} = 1.25 \times 10^{17}\) per second.

2. Power delivered \(= VI = 25000 \times 0.02 = 500\) W.

   X-ray power \(= 1\%\) of \(500 = 5\) W. Heat \(= 495\) J each second.

1. \(K_\alpha\): an electron from the L shell fills the K vacancy. \(K_\beta\): an electron from the M shell fills it.

2. The M shell is higher than the L shell, so the M\(\to\)K jump releases more energy. \(K_\beta\) has the higher energy and the shorter wavelength.

3. \(L_\alpha\) is an M\(\to\)L jump. Both shells are far from the nucleus and close in energy, so the photon energy is small and the wavelength long.

1. \(K_\alpha\) (L\(\to\)K): \(20.0 - 2.6 = 17.4\) keV, \(\lambda = \dfrac{12400}{17400} \approx 0.71\) Å.

   \(K_\beta\) (M\(\to\)K): \(20.0 - 0.5 = 19.5\) keV, \(\lambda \approx 0.64\) Å.

   \(L_\alpha\) (M\(\to\)L): \(2.6 - 0.5 = 2.1\) keV, \(\lambda \approx 5.9\) Å.

2. To make a K vacancy the incoming electron must supply \(20\) keV, the binding energy of the K shell. So the tube voltage must be at least \(20\) kV.

\(\nu \propto (Z - 1)^2\), so \(\lambda \propto \dfrac{1}{(Z - 1)^2}\).

\(\lambda_{Mo} = 1.54 \times \left(\dfrac{29 - 1}{42 - 1}\right)^2 = 1.54 \times \left(\dfrac{28}{41}\right)^2 = 1.54 \times 0.466 \approx 0.72\) Å.

1. The line meets the axis at \(Z = 1\). The inner electron feels the nuclear charge reduced by the one remaining K electron, an effective charge \((Z - 1)e\). A hypothetical element with \(Z = 1\) would give zero frequency.

2. At \(Z = 29\), \(\sqrt{\nu} = 1.39 \times 10^{9}\) Hz\(^{1/2}\). \(a = \dfrac{1.39 \times 10^{9}}{29 - 1} \approx 5.0 \times 10^{7}\) Hz\(^{1/2}\).

3. \(Z - 1 = \dfrac{1.64 \times 10^{9}}{4.97 \times 10^{7}} = 33\), so \(Z = 34\) (selenium).

1. For a hydrogen-like atom \(\nu = Rc\,Z_{eff}^2\left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right) = \dfrac{3}{4}Rc\,Z_{eff}^2\). With \(Z_{eff} = Z - 1\): \(\nu = \dfrac{3}{4}Rc\,(Z - 1)^2\).

2. \(\sqrt{\nu} = \sqrt{\dfrac{3Rc}{4}}\,(Z - 1)\), so \(a = \sqrt{0.75 \times 3.29 \times 10^{15}} = \sqrt{2.47 \times 10^{15}} \approx 4.97 \times 10^{7}\) Hz\(^{1/2}\). This matches the graph.

\(E = \dfrac{3}{4} \times 13.6\,(Z - 1)^2 = 10.2\,(Z - 1)^2\) eV.

1. Cu: \(10.2 \times 28^2 = 10.2 \times 784 \approx 8.0\) keV. \(\lambda = \dfrac{12400}{8000} \approx 1.55\) Å (measured: \(1.54\) Å).

2. Mo: \(10.2 \times 41^2 = 10.2 \times 1681 \approx 17.1\) keV. \(\lambda \approx 0.72\) Å (measured: \(0.71\) Å).

1. When one K electron is knocked out, one K electron remains. It screens the nucleus, so the electron that jumps in feels a charge of about \((Z - 1)e\), not \(Ze\).

2. The characteristic photon energy is the difference between two shell energies of the target atom. These energies are fixed by the atom. The tube voltage only decides whether the electrons have enough energy to create the vacancy.

\(E = \dfrac{12400}{0.76} \approx 16300\) eV. Then \((Z - 1)^2 = \dfrac{16300}{10.2} \approx 1600\), so \(Z - 1 = 40\) and \(Z = 41\) (niobium).

X-rays reflected from successive parallel planes of atoms reinforce when \(2d\sin\theta = n\lambda\).

\(d\) is the distance between neighbouring planes. \(\theta\) is the glancing angle, measured between the ray and the plane (not the normal). \(n = 1, 2, 3, \ldots\) is the order of the maximum.

1. \(\sin\theta = \dfrac{n\lambda}{2d} = \dfrac{n \times 1.0}{4.0} = 0.25n\).

   \(n = 1\): \(\sin\theta = 0.25\), \(\theta = 14.5^\circ\). \(n = 2\): \(\sin\theta = 0.5\), \(\theta = 30^\circ\).

2. \(\sin\theta \le 1\) needs \(0.25n \le 1\), so \(n \le 4\). The fourth order is at \(\theta = 90^\circ\), the highest possible.

1. Drop perpendiculars from the reflection point of ray 1 onto ray 2, before and after reflection. Each perpendicular cuts off an extra length \(d\sin\theta\) on ray 2. The total extra path is \(2d\sin\theta\).

2. The rays reinforce when the extra path is a whole number of wavelengths: \(2d\sin\theta = n\lambda\). That is Bragg's law.

1. \(d = \dfrac{n\lambda}{2\sin\theta} = \dfrac{1.54}{2 \times 0.5} = 1.54\) Å.

2. \(\sin\theta = \dfrac{\lambda}{2d} = \dfrac{1.41}{5.64} = 0.25\), so \(\theta = 14.5^\circ\).

1. The first peak is at \(14.5^\circ\). \(\sin 14.5^\circ = 0.25 = \dfrac{\lambda}{2d}\).

2. \(n = 2\): \(\sin\theta = 2 \times 0.25 = 0.5\), \(\theta = 30^\circ\), where the second peak is drawn.

3. \(n = 4\): \(\sin\theta = 4 \times 0.25 = 1\), \(\theta = 90^\circ\). It is just possible, with the rays perpendicular to the planes.

\(\sin\theta = \dfrac{n\lambda}{2d} \le 1\), so \(n \le \dfrac{2d}{\lambda}\). The highest order is the largest whole number that fits.

If \(\lambda > 2d\), even \(n = 1\) would need \(\sin\theta > 1\), which is impossible. Long waves cannot be diffracted by the crystal; this is why visible light shows no Bragg reflection.

1. \(\lambda = 2d\sin\theta = 2 \times 1.0 \times 0.5 = 1.0\) Å.

2. The tube can produce \(1.0\) Å only if \(\lambda_{\min} \le 1.0\) Å: \(V \ge \dfrac{12400}{1.0} = 12400\) V \(= 12.4\) kV.

1. \(E = \dfrac{12400}{0.5} = 24800\) eV \(= 24.8\) keV.

   \(p = \dfrac{h}{\lambda} = \dfrac{6.6 \times 10^{-34}}{0.5 \times 10^{-10}} = 1.3 \times 10^{-23}\) kg m/s.

2. \(E \propto \dfrac{1}{\lambda}\). \(\dfrac{500 \text{ nm}}{0.05 \text{ nm}} = 10^{4}\). The X-ray photon carries ten thousand times more energy.

1. \(40\) kV: \(\lambda_{\min} = \dfrac{12400}{40000} = 0.31\) Å. \(15\) kV: \(\lambda_{\min} = \dfrac{12400}{15000} \approx 0.83\) Å.

2. Only at \(40\) kV. The \(K_\alpha\) line needs a K-shell vacancy, which needs \(20\) keV electrons. At \(15\) kV the electrons cannot remove a K electron, so only the continuous spectrum appears, and it stops at \(0.83\) Å anyway.