Question 14.1: Complete the equations for each of the following nuclear dec......
Complete the equations for each of the following nuclear decay processes.
\begin{aligned}& { }_{84}^{210} Po \rightarrow{ }_{82}^{206} Pb +? \\& { }_{90}^{230} Th \rightarrow ?+{ }_2^4 He\end{aligned}
Strategy Nuclear equations must be balanced with respect to both total mass and charge. As long as only one particle is missing from each equation, we can use these two criteria to determine its identity.
Consider the first reaction. Looking at the two isotopes shown, the difference in atomic number is 84-82=2, and the difference in mass number is 210 – 206=4. That means that the missing particle must have a mass number of four and an atomic number of two, making it an alpha particle. The completed equation is
{ }_{84}^{210} Po \rightarrow{ }_{82}^{206} Pb +{ }_2^4 He
Next consider the second reaction. Again, looking at the two species shown, the difference in atomic number is 90-2=88 and the difference in mass number is 230-4=226. The fact that the atomic number is 88 tells us the missing isotope is radium, Ra, and the mass number tells us it must be radium-226. This lets us complete the equation:
{ }_{90}^{230} Th \rightarrow{ }_{88}^{226} Ra +{ }_2^4 He
Discussion This type of problem for alpha decay is normally quite straightforward. Later in the chapter, we will find nuclear reactions where, for example, several neutrons are produced as products. That type of problem is slightly more complicated because the stoichiometry changes from the simple one-to-one ratios seen here.
Check Your Understanding Identify the element and isotope of the missing reactant in the following nuclear equation.
? \rightarrow{ }_{82}^{205} Pb +{ }_2^4 He