Atomic nuclei and the stability of isotopes

The potassium in bananas contains a small fraction of the isotope potassium-40, whose nucleus has atomic number $Z = 19$ and mass number $A = 40$.

1. a) State how many protons and how many neutrons this nucleus has.
2. b) Write its symbol using ${}^{A}_{Z}\text{X}$ notation.
3. c) How many electrons does a neutral atom of potassium-40 have? Justify your answer.

Carbon has three natural isotopes: carbon-12, carbon-13 and carbon-14, all with $Z = 6$.

1. a) Explain what it means for two nuclei to be isotopes.
2. b) Calculate the number of neutrons in the nucleus of each of the three isotopes.
3. c) Do all three isotopes have the same chemical properties? Explain your answer by considering their electrons.

Consider these five nuclides: ${}^{14}_{6}\text{C}$, ${}^{14}_{7}\text{N}$, ${}^{12}_{6}\text{C}$, ${}^{16}_{8}\text{O}$ and ${}^{15}_{7}\text{N}$.

1. a) Calculate the number of protons and neutrons in each one.
2. b) State which pairs are isotopes (same $Z$) and which pair is isobars (same $A$).
3. c) Identify two nuclides that have the same number of neutrons (isotones).

The atomic mass unit (u) is defined as one twelfth of the mass of a carbon-12 atom, and is equivalent to $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$.

1. a) The mass of the oxygen-16 nucleus is about $16\ \text{u}$: express it in kilograms, in scientific notation.
2. b) Explain why in nuclear physics it is more convenient to work in units of u than in kilograms.

The radius of an atom is on the order of $10^{-10}\ \text{m}$, while that of its nucleus is on the order of $10^{-15}\ \text{m}$, as shown in the figure.

1. a) Calculate how many times larger the radius of the atom is than that of the nucleus.
2. b) The volume of a sphere is proportional to the cube of the radius: estimate how many times larger the volume of the atom is than that of the nucleus.
3. c) In light of the result, explain why it is said that the atom is mostly empty space.

The radius of a nucleus can be estimated using the expression $R = R_0 \times A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}\ \text{m}$ and $A$ is the mass number. For iron-56 take $A^{1/3} \approx 3.8$ and a nuclear mass of $9.3 \times 10^{-26}\ \text{kg}$. The volume of a sphere is $V = \frac{4}{3} \times \pi \times R^3$, with $\pi \approx 3.14$.

1. a) Calculate the radius of the iron-56 nucleus.
2. b) Calculate its volume and density, in $\text{kg/m}^3$.
3. c) The matter of a neutron star has a density similar to nuclear density: calculate the mass that a teaspoon of $5\ \text{cm}^3$ of this matter would have and comment on the result.

In a helium nucleus, two protons are confined in a region about $10^{-15}\ \text{m}$ in diameter. As both have positive charge, they repel each other electrically with considerable force; however, the nucleus does not break apart.

1. a) Explain what the strong nuclear force is and why it manages to keep the nucleus bound together.
2. b) Explain why, even with that force existing, nuclei with too many protons end up being unstable.

Of the nearly $3000$ known nuclides, barely $250$ are stable; the rest end up decaying sooner or later.

1. a) Explain what it means for a nucleus to be stable and what happens to an unstable nucleus.
2. b) In light stable nuclei, the number of neutrons $N$ and the number of protons $Z$ are almost equal: state the approximate value of the ratio $\frac{N}{Z}$ in these nuclei.

Two protons inside a nucleus are separated by a distance of $10^{-15}\ \text{m}$, as shown in the figure. Data: Coulomb constant $K = 9 \times 10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$ and proton charge $e = 1.6 \times 10^{-19}\ \text{C}$.

1. a) Calculate the repulsive electrostatic force between them using Coulomb's law, $F = K \times \frac{e^2}{r^2}$.
2. b) The value obtained is huge for a force between particles: explain what conclusion can be drawn about the strength of the strong nuclear force that holds them together.

The $\frac{N}{Z}$ ratio of stable nuclei is approximately $1$ for light nuclei and increases to about $1.5$ for the heaviest ones, as shown in the figure.

Calculate the $\frac{N}{Z}$ ratio for each of these nuclides and predict, by comparing with this trend, whether they are expected to be stable or unstable:

1. a) ${}^{12}_{6}\text{C}$
2. b) ${}^{14}_{6}\text{C}$
3. c) ${}^{206}_{82}\text{Pb}$
4. d) ${}^{238}_{92}\text{U}$

Oxygen has only three stable isotopes: ${}^{16}_{8}\text{O}$, ${}^{17}_{8}\text{O}$ and ${}^{18}_{8}\text{O}$. Isotopes with a neutron excess tend to decay by beta-minus emission ($\beta^{-}$: a neutron converts into a proton), and those with a neutron deficit, by beta-plus emission ($\beta^{+}$: a proton converts into a neutron).

1. a) Calculate the $\frac{N}{Z}$ ratio of ${}^{14}_{8}\text{O}$ and ${}^{21}_{8}\text{O}$ and compare it with that of ${}^{16}_{8}\text{O}$.
2. b) Predict which type of decay each of these two unstable isotopes will undergo, justifying your answer.
3. c) Write the equation for each process and identify the daughter nuclide formed in each case.

Precision measurements show that the mass of a helium-4 nucleus is less than the sum of the masses of its two protons and two neutrons measured separately.

1. a) Explain what the mass defect of a nucleus is.
2. b) What has the "missing" mass been transformed into? Mention Einstein's relation $E = \Delta m \times c^2$ in your explanation.

Data: mass of proton $m_p = 1.0073\ \text{u}$, mass of neutron $m_n = 1.0087\ \text{u}$, mass of helium-4 nucleus $m_{\alpha} = 4.0015\ \text{u}$ and $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$.

1. a) Calculate the sum of the masses of the two protons and two neutrons that form the helium-4 nucleus, measured separately.
2. b) Calculate the mass defect $\Delta m$ of the nucleus, in u units.
3. c) Express this mass defect in kilograms, in scientific notation.

The mass defect of the helium-4 nucleus is $\Delta m = 0.0305\ \text{u}$, equivalent to $5.06 \times 10^{-29}\ \text{kg}$. Data: $c = 3 \times 10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$.

1. a) Calculate the binding energy of the nucleus using Einstein's relation, $E = \Delta m \times c^2$, expressed in joules.
2. b) Convert the result to $\text{MeV}$ (note that $1\ \text{MeV} = 10^{6}\ \text{eV}$).
3. c) Calculate the binding energy per nucleon of helium-4.

In nuclear physics, the mass–energy equivalence $1\ \text{u} \times c^2 = 931.5\ \text{MeV}$ is used.

1. a) The mass defect of deuterium (hydrogen-2 nucleus) is $\Delta m = 0.0024\ \text{u}$: calculate its binding energy in $\text{MeV}$.
2. b) Calculate its binding energy per nucleon and compare it with that of iron-56 ($8.8\ \text{MeV}$ per nucleon): explain which of the two nuclei is more stable.

The total binding energy of tritium (hydrogen-3 nucleus) is $8.48\ \text{MeV}$. Data: $1\ \text{u} \times c^2 = 931.5\ \text{MeV}$, $m_p = 1.0073\ \text{u}$ and $m_n = 1.0087\ \text{u}$.

1. a) Calculate the mass defect of tritium, in u units.
2. b) Given that tritium has one proton and two neutrons, calculate the mass of its nucleus.
3. c) Calculate its binding energy per nucleon and, comparing it with that of helium-4 ($7.1\ \text{MeV}$), reason why the fusion of deuterium and tritium to form helium releases energy.

Shortly after the Big Bang, protons and neutrons in the early universe joined together to form deuterium, and in each union a gamma photon was emitted. Data: $m_p = 1.0073\ \text{u}$, $m_n = 1.0087\ \text{u}$, mass of deuterium nucleus $m_d = 2.0136\ \text{u}$ and $1\ \text{u} \times c^2 = 931.5\ \text{MeV}$.

1. a) Calculate the mass defect of the process in which a proton and a neutron form a deuterium nucleus.
2. b) Calculate the energy released in the formation of each deuterium nucleus, in $\text{MeV}$.
3. c) Where does this energy go? Explain the role of the emitted gamma photon.

All the electricity consumed in Spain in a year is of the order of $9 \times 10^{17}\ \text{J}$. Data: $c = 3 \times 10^{8}\ \text{m/s}$.

1. a) Suppose that this energy were obtained by converting mass directly into energy, with an efficiency of $100\ \%$: calculate, using $E = \Delta m \times c^2$, the mass that would have to be converted.
2. b) Comment on the result: does it seem like a lot or a little mass to supply electricity to an entire country for a year?
3. c) A student states: "in nuclear reactions mass is conserved, just as in chemical reactions". Correct their statement and explain what actually is conserved in a nuclear reaction.

The binding energy per nucleon is the average energy required to remove a nucleon from a nucleus.

1. a) Define the binding energy of a nucleus.
2. b) The binding energy per nucleon is maximum in intermediate-mass nuclei, such as iron, and lower in very light and very heavy nuclei. Explain which nuclei will be the most stable.

In nuclide tables, the binding energy per nucleon of iron-56 is given as $8.8\ \text{MeV}$ and that of uranium-235 as $7.6\ \text{MeV}$, as shown in the figure.

1. a) Calculate the total binding energy of each of these two nuclei.
2. b) Explain why iron-56 is more stable than uranium-235 even though its total binding energy is smaller.
3. c) Relate the result to the fact that uranium-235 releases energy when undergoing fission.

Data: mass of proton $m_p = 1.0073\ \text{u}$, mass of neutron $m_n = 1.0087\ \text{u}$, mass of iron-56 nucleus $m(\text{Fe}) = 55.9207\ \text{u}$ and $1\ \text{u} \times c^2 = 931.5\ \text{MeV}$.

1. a) Iron-56 has $Z = 26$: state how many protons and how many neutrons its nucleus contains.
2. b) Calculate its mass defect, in u units.
3. c) Calculate its total binding energy and its binding energy per nucleon, in $\text{MeV}$.
4. d) Given that the binding energy per nucleon of carbon-12 is $7.7\ \text{MeV}$, reason which of the two nuclei is more stable and which process, fusion or fission, could release energy starting from each one.

The curve of binding energy per nucleon against mass number $A$ increases rapidly from the lightest nuclei, reaches a maximum around iron-56 ($8.8\ \text{MeV}$ per nucleon) and then decreases gently to about $7.6\ \text{MeV}$ in uranium-235.

1. a) Sketch this curve in your notebook, marking the maximum and the regions of light and heavy nuclei.
2. b) Explain, from the shape of the curve, why the fusion of light nuclei releases energy.
3. c) Explain, using the same argument, why the fission of very heavy nuclei also releases energy.
4. d) Reason why energy cannot be obtained by either fusing or fissioning iron nuclei.

You can do it on paper and send a photo with your answer.

When a massive star exhausts the fuel in its core, it ends up producing iron-56, a nuclide lying practically at the maximum of the binding energy per nucleon.

1. a) Explain, using the binding energy per nucleon curve, why fusing iron does not release energy, but instead consumes it.
2. b) Reason what consequence this has for the star, which no longer has an energy source in its core.
3. c) Elements heavier than iron are mainly formed by neutron capture during supernova explosions: explain why these elements are much less abundant in the universe than iron.