Writing Nuclear Equations for Radioactive Decay Processes Write nuclear equations to represent (a) α-particle emission by ^222Rn and (b) radioactive decay of bismuth-215 to polonium-215. Analyze In part (a) we can identify two of the species involved in this process from the information given and

Question

Writing Nuclear Equations for Radioactive Decay Processes

Write nuclear equations to represent (a) α-particle emission by [latex]_{}^{222} extrm{Rn}[/latex] and (b) radioactive decay of bismuth-215 to polonium-215.

Analyze
In part (a) we can identify two of the species involved in this process from the information given and we can write an incomplete equation. The unknown element in the incomplete equation is identified by determining the atomic number (Z) and mass number (A) that will balance the incomplete equation. Part (b) is done in a way very similar to that in part (a).

Accepted Answer

(a) Since the [latex]_{86}^{222} extrm{Rn}[/latex] nucleus ejects an α particle, [latex]_{2}^{4} extrm{He}[/latex], as shown in the following incomplete nuclear equation.

[latex]{ }_{86}^{222} ext{Rn} \longrightarrow ext{ ?} +{ }_2^4 ext{He}[/latex]

Because the ejected α particle contains two protons, the unknown product must contain two fewer protons than [latex]_{86}^{222} extrm{Rn: } Z = 86 - 2 = 84[/latex]. This atomic number identifies the element as polonium, [latex]_{84}^{} extrm{Po}[/latex]. The mass number (A) of the product can be obtained by subtracting the mass number of the α particle from that of the radon isotope: A = 222 - 4 = 218. The completed nuclear equation is

[latex]{ }_{86}^{222} ext{Rn} \longrightarrow{ }_{84}^{218} ext{Po} +{ }_2^4 ext{He}[/latex]

(b) The atomic number of bismuth is 83 and that of polonium is 84. We can approach this problem as we did part (a).

[latex]{ }_{83}^{215} ext{Bi} \longrightarrow{ }_{84}^{215} ext{Po} + ?[/latex]

There is no change in mass number, so the particle has a zero mass number. Its atomic number is Z = 83 - 84 = -1. Only a [latex]_{-1}^{0}\beta [/latex] particle fits these parameters: Beta (−) decay is the only type of emission leading to an increase of one unit in atomic number without a change in the mass number.

[latex]{ }_{83}^{215} ext{Bi} \longrightarrow{ }_{84}^{215} ext{Po} +{ }_{-1}^0 \beta[/latex]

Assess
It is interesting that when starting from different elements, we can arrive at the same element through different types of particle emission. Note that even though we came to the same decayed element, we ended with different isotopes of that element.

Question 25.1: Writing Nuclear Equations for Radioactive Decay Processes Wr...

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