Question

nuclear decay series
the figure shown maps the radioactive decay of uranium - 238. use the figure to answer the following questions.

1. how many alpha particles are produced when one atom of u - 238 decays into pb - 206?
2. how many beta particles?
3. what is the final product of this decay series?

Explanation:

Step1: Recall alpha - decay rule

In alpha - decay, the mass number decreases by 4 and the atomic number decreases by 2. Uranium - 238 has a mass number of 238 and lead - 206 has a mass number of 206. The change in mass number is $\Delta A=238 - 206=32$. Since each alpha - decay changes the mass number by 4, the number of alpha - decays $n_{\alpha}=\frac{238 - 206}{4}=8$.

Step2: Recall beta - decay rule

In beta - decay, the mass number remains the same and the atomic number increases by 1. Uranium has an atomic number of 92 and lead has an atomic number of 82. After 8 alpha - decays, the atomic number would decrease by $8\times2 = 16$ from 92 to $92-16 = 76$. But the final atomic number of lead is 82. So the number of beta - decays $n_{\beta}=82-(92 - 16)=6$.

Step3: Identify final product

The problem states that the decay of uranium - 238 ends up at lead - 206.

Answer:

1. 8
2. 6
3. Pb - 206