Changing one nucleus into another
In a nuclear reaction, a particle strikes a nucleus and something new comes out. A light projectile \(a\) hits a target nucleus \(X\). A new nucleus \(Y\) is left behind and a light particle \(b\) flies off. The reaction is written \(a + X \to Y + b\), or in short \(X(a, b)Y\).
Alchemists had tried for centuries to turn one element into another. In 1919 Rutherford did it with alpha particles and nitrogen gas. He saw flashes on his screen much further away than any alpha particle could reach. They were protons, knocked out of nitrogen nuclei. In 1925 Patrick Blackett photographed 400,000 alpha tracks in a cloud chamber. Eight of them ended in a nitrogen nucleus that turned into oxygen-17 and sent out a proton.
Rutherford's reaction was \({}^{4}_{2}\text{He} + {}^{14}_{7}\text{N} \to {}^{17}_{8}\text{O} + {}^{1}_{1}\text{H}\), or \({}^{14}\text{N}(\alpha, p){}^{17}\text{O}\). Add the numbers on each side. The mass numbers give \(4 + 14 = 17 + 1 = 18\). The atomic numbers give \(2 + 7 = 8 + 1 = 9\). Both totals are the same before and after.
Every nuclear reaction conserves the total charge, so the sum of the atomic numbers is the same before and after. It conserves the total number of nucleons, the sum of the mass numbers. It conserves the total energy, counting rest energy \(mc^2\) as well as kinetic energy. It also conserves the total momentum and angular momentum.
Notice what is missing from the list: mass. The total mass before and after is slightly different, because rest energy and kinetic energy can turn into each other. That small change of mass is the whole subject of the next topic.
The first two laws let you fill in a missing particle. In 1934 Irène Curie and Frédéric Joliot fired alpha particles at aluminium, \({}^{27}_{13}\text{Al}(\alpha, n)\,?\). The nucleon count gives \(27 + 4 - 1 = 30\), and the charge gives \(13 + 2 - 0 = 15\). So the product is \({}^{30}_{15}\text{P}\), an isotope of phosphorus never seen before. It was radioactive: the first radioactive nucleus made by people.
In the reaction \({}^{10}_{5}\text{B} + {}^{1}_{0}n \to {}^{7}_{3}\text{Li} + ?\), the missing particle is
The Q value
Conservation of energy, with rest energy included, reads \(m_ic^2 + K_i = m_fc^2 + K_f\). Here \(m_i\) and \(K_i\) are the total mass and kinetic energy before, and \(m_f\) and \(K_f\) after. Rearranged, the kinetic energy gained equals the rest energy lost. That amount is the Q value of the reaction.
The sign of \(Q\) sorts reactions into two kinds. If \(Q > 0\), the products are lighter than the reactants, and the lost mass appears as extra kinetic energy: the reaction is exothermic. If \(Q < 0\), the products are heavier, and kinetic energy must be turned into mass: the reaction is endothermic. The same \(Q\) also comes from binding energies, \(Q = B_{\text{after}} - B_{\text{before}}\), as in the last lesson.
Find \(Q\) for \({}^{14}\text{N} + {}^{4}\text{He} \to {}^{17}\text{O} + {}^{1}\text{H}\). Atomic masses: \({}^{14}\text{N}\) \(14.003074\,\text{u}\), \({}^{4}\text{He}\) \(4.002603\,\text{u}\), \({}^{17}\text{O}\) \(16.999132\,\text{u}\), \({}^{1}\text{H}\) \(1.007825\,\text{u}\). Is the reaction exothermic or endothermic?
Add the masses on each side. Atomic masses are fine: both sides hold 9 electrons.
\(Q = (m_{\text{initial}} - m_{\text{final}}) \times 931.5\,\text{MeV/u}\).
Initial: \(14.003074 + 4.002603 = 18.005677\,\text{u}\). Final: \(16.999132 + 1.007825 = 18.006957\,\text{u}\). \(\Delta m = -0.001280\,\text{u}\), so \(Q = -0.001280 \times 931.5 = -1.19\,\text{MeV}\).
\(Q = -1.19\,\text{MeV}\): the reaction is endothermic. The alpha particle's kinetic energy must provide at least 1.19 MeV of new rest energy. Values used: \(1\,\text{u} = 931.5\,\text{MeV}/c^2\) and the four atomic masses above.
Other reactions release energy. In 1932 John Cockcroft and Ernest Walton fired protons from their accelerator at lithium: \({}^{7}\text{Li} + p \to 2\,{}^{4}\text{He}\). The masses, \(7.016003 + 1.007825 = 8.023828\,\text{u}\) before and \(2 \times 4.002603 = 8.005206\,\text{u}\) after, give \(Q = 0.018622 \times 931.5 = 17.35\,\text{MeV}\). The two alpha particles flew apart with far more energy than the protons brought in.
The second form, \(Q = K_f - K_i\), needs no masses at all. Suppose a proton of \(1\,\text{MeV}\) strikes a nucleus at rest, and the products leave with \(4\,\text{MeV}\) in all. Then \(Q = 4 - 1 = 3\,\text{MeV}\). Turned round, it predicts the products. A deuteron of \(1\,\text{MeV}\) striking a deuteron at rest, with \(Q = 3.27\,\text{MeV}\), gives products with \(4.27\,\text{MeV}\) in all.
Find the Q value of \({}^{2}\text{H} + {}^{2}\text{H} \to {}^{3}\text{He} + n\), in MeV. Atomic masses: \({}^{2}\text{H}\) \(2.014102\,\text{u}\), \({}^{3}\text{He}\) \(3.016029\,\text{u}\), \(n\) \(1.008665\,\text{u}\). Take \(1\,\text{u} = 931.5\,\text{MeV}/c^2\).
A reaction has \(Q = -2\,\text{MeV}\). What does the negative sign tell you?
How the products share the energy
When the reactants have almost no momentum, the products must fly apart with equal and opposite momenta. A slow neutron caught by boron-10 is such a case: \(n + {}^{10}\text{B} \to {}^{7}\text{Li} + \alpha\), with \(Q = 2.79\,\text{MeV}\). Each product has momentum \(p\), and kinetic energy \(K = p^2/2m\). With the same \(p\), the lighter product takes the larger share.
For boron, \(K_\alpha/K_{\text{Li}} = m_{\text{Li}}/m_\alpha = 7/4\). The two shares add up to \(2.79\,\text{MeV}\), so \(K_\alpha = \tfrac{7}{11} \times 2.79 = 1.78\,\text{MeV}\) and \(K_{\text{Li}} = \tfrac{4}{11} \times 2.79 = 1.01\,\text{MeV}\). The mass numbers are close enough to the masses for this. When the reactants do bring momentum, as a fast deuteron does, the products share that momentum too, and their total is not zero.
Deuterium and tritium, both nearly at rest, fuse: \({}^{2}\text{H} + {}^{3}\text{H} \to {}^{4}\text{He} + n\), releasing \(17.6\,\text{MeV}\). What kinetic energy does the neutron carry, in MeV?
Why |Q| is not enough
Rutherford's reaction needs \(1.19\,\text{MeV}\). Will an alpha particle of exactly \(1.19\,\text{MeV}\) make it happen? No. The alpha particle brings momentum, and momentum is conserved. After the collision the products must still carry that momentum, so they cannot be at rest. Part of the kinetic energy must stay as motion, and only the rest can pay for the new mass.
The collision with the least wasted energy is the one in which the products move off together, at the velocity of the centre of mass. This is the same argument as the least energy lost when two bodies stick, from the chapter on Collisions. The chapter on the Bohr atom used it for exciting an atom by a collision.
\[ p = \sqrt{2mK} \]
Why. the projectile of mass \(m\) and kinetic energy \(K\) brings momentum \(p\); the target of mass \(M\) is at rest.
\[ K' = \dfrac{p^2}{2(m + M)} = \dfrac{m}{m + M}\,K \]
Why. at threshold the products move together with the same momentum \(p\) and total mass about \(m + M\).
\[ K - K' = \dfrac{M}{m + M}\,K \]
Why. this part of \(K\) is free to be turned into rest energy; the part \(K'\) is locked in the motion of the centre of mass.
\[ \dfrac{M}{m + M}\,K_{th} = |Q| \;\Rightarrow\; K_{th} = |Q|\left(1 + \dfrac{m}{M}\right) \]
Read it. the reaction just becomes possible when the free part equals \(|Q|\).
For Rutherford's reaction, \(m = 4\) and \(M = 14\). So \(K_{th} = 1.19 \times 18/14 = 1.53\,\text{MeV}\), as the figure above shows. An alpha particle of \(1.40\,\text{MeV}\) has \(1.09\,\text{MeV}\) available, which is not enough. The threshold is smaller when the target is heavy compared with the projectile: then almost all of \(K\) is available.
Protons striking lithium-7 make neutrons: \({}^{7}\text{Li}(p, n){}^{7}\text{Be}\). Atomic masses: \({}^{7}\text{Li}\) \(7.016003\,\text{u}\), \({}^{1}\text{H}\) \(1.007825\,\text{u}\), \({}^{7}\text{Be}\) \(7.016929\,\text{u}\), \(n\) \(1.008665\,\text{u}\). Find (a) the Q value and (b) the threshold kinetic energy of the protons.
(a) Initial: \(7.016003 + 1.007825 = 8.023828\,\text{u}\). Final: \(7.016929 + 1.008665 = 8.025594\,\text{u}\). \(\Delta m = -0.001766\,\text{u}\), so \(Q = -1.645\,\text{MeV}\).
(b) \(K_{th} = 1.645 \times (1 + 1/7) = 1.645 \times 8/7 = 1.88\,\text{MeV}\). Laboratories use exactly this: below 1.88 MeV the target gives no neutrons, and just above it the neutrons appear.
A photon can also start an endothermic reaction. A gamma ray splits a deuteron, \(\gamma + {}^{2}\text{H} \to p + n\), with \(Q = -2.22\,\text{MeV}\). A photon has momentum \(E/c\), so the products must again keep some motion. The exact threshold is \(|Q|(1 + |Q|/2Mc^2)\), where \(Mc^2 = 1876\,\text{MeV}\) is the deuteron's rest energy. The recoil adds only \(1.3\,\text{keV}\), because the photon's momentum is tiny for its energy.
A positive \(Q\) does not always mean that no energy is needed. For \({}^{7}\text{Li} + p \to 2\alpha\), with \(Q = +17.35\,\text{MeV}\), there is no threshold from energy. Yet the proton and the lithium nucleus are both positive and repel. The proton must come within reach of the short-range nuclear force, so it must climb this Coulomb barrier first. A neutron has no charge and feels no barrier, which is why slow neutrons react so easily.
A reaction started by a neutron (mass 1) on carbon-12 at rest has \(Q = -5.0\,\text{MeV}\). Find the threshold kinetic energy of the neutron, in MeV.
An alpha particle with kinetic energy exactly \(1.19\,\text{MeV}\) strikes nitrogen-14 at rest. The reaction has \(Q = -1.19\,\text{MeV}\). Why does the reaction not happen?
- Taking \(Q\) as final mass minus initial mass.\(Q = (m_{\text{initial}} - m_{\text{final}})c^2\). Positive \(Q\) means energy released.
- Using nuclear masses on one side and atomic masses on the other.Use atomic masses throughout. The electrons balance because the charge does.
- Setting the threshold kinetic energy equal to \(|Q|\).The products must keep the momentum. \(K_{th} = |Q|(1 + m/M)\), always more than \(|Q|\).
- Giving the two products equal energies when they fly apart from rest.They have equal momenta. \(K = p^2/2m\), so the lighter product takes more.