Question

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

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

9223582207Pb +?24He+?1 0e\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 α\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 α\alpha-particle

2352074=7number of α-particle\begin{align*} \dfrac{235-207}{4} &= 7&\text{number of $\alpha$-particle}\\ \end{align*}

An α\alpha-particle has an atomic number of 2, we have 7 α\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 β\beta-particle has an atomic number of -1, thus to balance the reaction, we need 4 β\beta-particles

82+17=96exceeds atomic number of U-235 by 496+4(1)=92atomic number of U-235\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 α\alpha and β\beta decay is as follows:

9223582207Pb +724He+41 0e\begin{align*} ^{235}_{92}\text{U }\rightarrow^{207}_{82}\text{Pb } + 7 ^{4}_{2}\text{He} + 4 ^\text{ 0}_{-1}\text{e} \end{align*}

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