Question

Question #575fe

Answer 1

`""_ (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 `color(red)(x)` and `color(blue)(y)`. You can write

`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 `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.

`""_ (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, `bar(nu)_"e"`, so the complete nuclear equation would look like this

`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.