write the balanced nuclear equation to describe beta decay process for aluminum - 29. write the isotope name…

Explanation:

Step1: Recall beta - decay concept

In beta - decay, a neutron in the nucleus is converted into a proton, an electron (beta particle), and an antineutrino. The atomic number increases by 1 while the mass number remains the same. The atomic number of Aluminum (Al) is 13 and its mass number here is 29. The symbol for Aluminum - 29 is $^{29}{13}Al$. The symbol for a beta particle is $^{\ \ 0}{- 1}\beta$.

Step2: Write the nuclear equation

The general form of beta - decay is $^{A}{Z}X\rightarrow^{A}{Z + 1}Y+^{\ \ 0}{- 1}\beta+\bar{\nu}e$. For Aluminum - 29 ($^{29}{13}Al$), after beta - decay, the new element has an atomic number of $13+1 = 14$ (which is Silicon, Si) and a mass number of 29. The balanced nuclear equation is $^{29}{13}Al\rightarrow^{29}{14}Si+^{\ \ 0}{- 1}\beta+\bar{\nu}e$. We usually omit the antineutrino in simple nuclear equations, so the equation is $^{29}{13}Al\rightarrow^{29}{14}Si+^{\ \ 0}{- 1}\beta$.

Answer:

$^{29}{13}Al\rightarrow^{29}{14}Si+^{\ \ 0}_{- 1}\beta$; Silicon - 29