# Question #575fe

##### 1 Answer

#### Answer:

#### 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

#113 = 113 + color(blue)(y) -># conservation of mass

#color(white)(a)47 = color(white)(a)48 + color(red)(x) -># conservation of charge

This gets you

#113 = 113 + color(blue)(y) implies color(blue)(y) = 0#

and

#47 = 48 + color(red)(x) implies color(red)(x) = 47 - 48 = -1#

Therefore, your unknown particle has a mass number of *high-speed electron*, also called **beta particle**,

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

#""_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + ""_(-1)^(color(white)(aa)0)beta#

It's worth mentioning that beta minus decay also produces an *electron antineutrino*,

#color(green)(|bar(ul(color(white)(a/a)color(black)(""_ (color(white)(a)47)^113"Ag" -> ""_ (color(white)(a)48)^113"Cd" + ""_ (-1)^(color(white)(aa)0)beta + bar(nu)_"e")color(white)(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.