Question:In a natural decay series, how many$\mathit{\alpha }$and${\mathit{\beta }}^{\mathbf{-}}$emissions per atom of uranium-235 result in an atom of lead-207?

When a nuclide loses doubly ionised ${}_{\mathbf{2}}\mathbf{He}^{\mathbf{4}}$nuclide then it is called alpha-decay.

When it loses ${}_{\mathbf{-}\mathbf{1}}{\mathbf{e}}^{\mathbf{0}}$the atomic number of the element increases by one and it is called beta-decay.

We have a natural decay series of uranium-235 $\left({}_{92}{}^{235}\text{U}\right)$, with the final product being lead-207 $\left({}_{82}{}^{207}\text{Pb}\right)$.

The amount of alpha and beta emissions in this sequence must also be determined.

${}_2{}^4\text{He}$is the alpha particle.

${}_1{}^{0}e$for the beta particle

We know that atomic mass falls by 4 and atomic number decreases by 2, whereas atomic mass remains constant and atomic number increases by 1 in each alpha decay.

We'll start by calculating the number of alpha decays because atomic mass changes exclusively in alpha decay.

$\begin{array}{rcl}\text{No alpha decays}& =& \frac{\left(\text{atomic mass U}\right)-\left(\text{atomic mass Pb}\right)}{\text{atomic mass alpha particle}}\\ & =& \frac{235-207}{4}\\ & =& 7\end{array}$

Therefore, number of alpha decays is 7.

After 7 alpha decays of uranium-235, we must now determine the atomic number of a product.

$\begin{array}{rcl}\text{Z}& =& \text{atomic number of} \text{U}-7\times \text{atomic number alpha particle}\\ & =& 92-7\times 2\\ & =& 78\end{array}$

$\begin{array}{rcl}\text{No beta decays}& =& \frac{\text{Z}-\text{atomic number Pb}}{\text{atomic number beta particle}}\\ & =& \frac{78-82}{-1}\\ & =& 4\end{array}$

Therefore, the number of beta decay is 4.