# Question 575fe

Jun 23, 2016

$\text{_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + ""_ (-1)^(color(white)(aa)0)beta + bar(nu)_"e}$

#### Explanation:

Your strategy here will be to use the fact that in a nuclear reaction, mass and charge are conserved.

In your case, the nuclear reaction involves the decay of silver-113 to cadmium-113

""_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + ""_color(red)(x)^color(blue)(y)?

Your goal now is to find the values of $\textcolor{red}{x}$ and $\textcolor{b l u e}{y}$. You can write

$113 = 113 + \textcolor{b l u e}{y} \to$ conservation of mass

$\textcolor{w h i t e}{a} 47 = \textcolor{w h i t e}{a} 48 + \textcolor{red}{x} \to$ conservation of charge

This gets you

$113 = 113 + \textcolor{b l u e}{y} \implies \textcolor{b l u e}{y} = 0$

and

$47 = 48 + \textcolor{red}{x} \implies \textcolor{red}{x} = 47 - 48 = - 1$

Therefore, your unknown particle has a mass number of $0$ and a charge of $1 -$. At this point, you should recognize that your unknown particle is a high-speed electron, also called beta particle, ""_(-1)^(color(white)(aa)0)beta#.

The nuclear equation given to you describes the beta minus decay of silver-113.

${\text{_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + }}_{- 1}^{\textcolor{w h i t e}{a a} 0} \beta$

It's worth mentioning that beta minus decay also produces an electron antineutrino, ${\overline{\nu}}_{\text{e}}$, so the complete nuclear equation would look like this

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + ""_ (-1)^(color(white)(aa)0)beta + bar(nu)_"e}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As you can see, in beta minus decay, a neutron is converted into a proton and an electron and an antineutrino are emitted from the nucleus.