Question

# Uranium-235 undergoes a series of a-particle and $\beta$-particle productions to end up as lead-207. How many $\alpha$ particles and $\beta$ particles are produced in the complete decay series?

Solution

Verified
Step 1
1 of 2

Form a nuclear equation of uranium-235 decaying to lead-207 including the $\alpha$-particles, and $\beta$-particles

\begin{align*} ^{235}_{92}\text{U }\rightarrow^{207}_{82}\text{Pb } + ? ^{4}_{2}\text{He} +? ^\text{ 0}_{-1}\text{e} \end{align*}

We know that $\alpha$-particles can change the mass number, thus we balance the mass number first

Subtract the mass number of lead-207 from the mass number of uranium-235 and divide it by the mass number of helium which has the same mass number of an $\alpha$-particle

\begin{align*} \dfrac{235-207}{4} &= 7&\text{number of \alpha-particle}\\ \end{align*}

An $\alpha$-particle has an atomic number of 2, we have 7 $\alpha$-particles , thus we have a total atomic number of 14 added to the atomic number of lead-207, exceeds the atomic number of uranium-235 by 4

A $\beta$-particle has an atomic number of -1, thus to balance the reaction, we need 4 $\beta$-particles

\begin{align*} 82+17 &= 96&\text{exceeds atomic number of U-235 by 4}\\ 96 + 4(-1) &= 92&\text{atomic number of U-235}\\ \end{align*}

Thus, the balanced nuclear reaction of U-235 decaying to lead-207 by a series of $\alpha$ and $\beta$ decay is as follows:

\begin{align*} ^{235}_{92}\text{U }\rightarrow^{207}_{82}\text{Pb } + 7 ^{4}_{2}\text{He} + 4 ^\text{ 0}_{-1}\text{e} \end{align*}

## Recommended textbook solutions

#### Chemistry: The Molecular Nature of Matter and Change

7th EditionISBN: 9780073511177Patricia Amateis, Silberberg
3,832 solutions

#### Chemistry

9th EditionISBN: 9781133611097 (23 more)Steven S. Zumdahl, Susan A. Zumdahl
5,602 solutions

#### Chemistry

10th EditionISBN: 9781305957404 (2 more)Donald J. DeCoste, Steven S. Zumdahl, Susan A. Zumdahl
5,855 solutions

#### Chemistry, AP Edition

10th EditionISBN: 9781305957732Donald J. DeCoste, Steven S. Zumdahl, Susan A. Zumdahl
5,932 solutions