Question #575fe

1 Answer
Jun 23, 2016

Answer:

#""_ (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.

https://www.learner.org/courses/physics/visual/visual.html?shortname=carbon14decay