A Beta Particle May Be Spontaneously Emitted From

Solution

Radon has an atomic number of 86, so the parent isotope is represented as \(\ce{^{222}_{86}Rn}\). We represent the alpha particle as \(\ce{^{4}_{2}He}\) and utilize subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as an isotope of polonium, \(\mathrm{^{218}_{84}Po}\):

\[\ce{_{86}^{222}Rn\rightarrow \, _2^4He + \, _{84}^{218}Po} \nonumber\]

Solution

The parent isotope is \(\ce{^{12}_{five}B}\) while one of the products is an electron, \(\ce{^{0}_{-one}e}\). So that the mass and diminutive numbers take the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an atomic number of 6 is carbon. Thus, the consummate nuclear equation is as follows:

\[\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0e + \gamma} \nonumber\]

The girl isotope is \(\ce{^{12}_6 C}\).

Solution

a.

A
We know the identities of the reactant and one of the products (a β particle). We can therefore brainstorm past writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\): \[^{35}_{xvi}\textrm{Southward}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-i}\beta \nonumber\]

B
Considering 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 − (−ane) = 17. The element with
Z
= 17 is chlorine, and then the counterbalanced nuclear equation is as follows: \[^{35}_{sixteen}\textrm{South}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber\]

b.

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: \[^{201}_{fourscore}\textrm{Hg}+\,^{0}_{-1}\textrm eastward\rightarrow\,^{A}_{Z}\textrm{X} \nonumber\]

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 exist
Z
= lxxx + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus \[^{201}_{lxxx}\textrm{Hg}+\,^{0}_{-1}\textrm eastward\rightarrow\,^{201}_{79}\textrm{Au} \nonumber\]

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 \[^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+i}\beta \nonumber\]

B
The mass number of the second product is
A
= 30 − 0 = 30, and its atomic number is
Z
= xv − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \[^{30}_{xv}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+ane}\beta \nonumber\]

Solution

1. This nuclide has a neutron-to-proton ratio of but 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that take 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 example, both are observed, with positron emission occurring almost 86% of the time and electron capture about xiv% of the time.
2. Nuclei with
   Z
   > 83 are as well heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the diminutive number. Thus \(^{242}_{94}\textrm{Pu}\) is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.iv, which is very high for a lite chemical 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 diminutive number with no change in the mass number. Nosotros therefore predict that \(^{12}_{v}\textrm{B}\) will undergo beta decay.
4. 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 past either or both of these two processes. In fact, it decays past both spontaneous fission and alpha emission, in a 97:3 ratio.