# 19.5: Nuclear Fission

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Only very massive nuclei with high neutron-to-proton ratios can undergo **spontaneous fission**, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with *Z* ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(^{254}_{98}\textrm{Cf}\), which gives a distribution of fission products; one possible set of products is shown in the following equation:

\[^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n}\label{5.2.16}\]

Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.

Example \(\PageIndex{1}\)

Write a balanced nuclear equation to describe each reaction.

- the beta decay of \(^{35}_{16}\textrm{S}\)
- the decay of \(^{201}_{80}\textrm{Hg}\) by electron capture
- the decay of \(^{30}_{15}\textrm{P}\) by positron emission

**Given:** radioactive nuclide and mode of decay

**Asked for:** balanced nuclear equation

**Strategy:**

**A** Identify the reactants and the products from the information given.

**B** Use the values of *A* and *Z* to identify any missing components needed to balance the equation.

**Solution:**

a.

**A** We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\):

**B** Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of *A* = 35 − 0 = 35 and an atomic number of *Z* = 16 − (−1) = 17. The element with *Z* = 17 is chlorine, so the balanced nuclear equation is as follows:

**A** We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows:

\(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}\)

**B** Both protons and neutrons are conserved, so the mass number of the product must be *A* = 201 + 0 = 201, and the atomic number of the product must be *Z* = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus

\(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}\)

c.**A** As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore

\(^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta\)

**B** The mass number of the second product is *A* = 30 − 0 = 30, and its atomic number is *Z *= 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows:

\(^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta\)

Exercise \(\PageIndex{1}\)

Write a balanced nuclear equation to describe each reaction.

- \(^{11}_{6}\textrm{C}\) by positron emission
- the beta decay of molybdenum-99
- the emission of an α particle followed by gamma emission from \(^{185}_{74}\textrm{W}\)

**Answer**

- \(^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta\)
- \(^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta\)
- \(^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma\)

Example \(\PageIndex{2}\)

Predict the kind of nuclear change each unstable nuclide undergoes when it decays.

- \(^{45}_{22}\textrm{Ti}\)
- \(^{242}_{94}\textrm{Pu}\)
- \(^{12}_{5}\textrm{B}\)
- \(^{256}_{100}\textrm{Fm}\)

**Given:** nuclide

**Asked for:** type of nuclear decay

**Strategy:**

Based on the neutron-to-proton ratio and the value of *Z*, predict the type of nuclear decay reaction that will produce a more stable nuclide.

**Solution:**

- This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
- Nuclei with
*Z*> 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus \(^{242}_{94}\textrm{Pu}\) is expected to decay by alpha emission. - This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that \(^{12}_{5}\textrm{B}\) will undergo beta decay.
- This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with
*A*≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that \(^{256}_{100}\textrm{Fm}\) will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.

Exercise \(\PageIndex{2}\)

Predict the kind of nuclear change each unstable nuclide undergoes when it decays.

- \(^{32}_{14}\textrm{Si}\)
- \(^{43}_{21}\textrm{Sc}\)
- \(^{231}_{91}\textrm{Pa}\)

**Answer**

- beta decay
- positron emission or electron capture
- alpha decay