Question

Complete the nuclear equation?

My guess is:
1. 98/44 Ru
2. 52/24 Cr

Thanks

Answer 1

Both of your guesses are correct.

Explanation:

1. Decay of `bb(""_45^98"Rh")`

`""_45^98"Rh" -> ""_1^0"e" + ?`

Positron (`""_1^0"e"`) is formed in the reaction. Therefore it’s a `β` positive decay.

In a `β` positive decay proton turns into neutron. Therefore the resultant atom should have

`"Protons = 45 – 1 = 44"`
`"Neutrons = (98 – 45) + 1 = 54"`

An atom with `44` protons is `"Ru"`

So, `β^+` decay reaction of `""_45^98"Rh"` is

`""_45^98"Rh" -> ""_44^98"Ru" + ""_1^0"e" + ν`

——————

2. Decay of `bb(""_23^52"V")`

`""_23^52"V" -> ""_"-1"^0"e" + ?`

`β` particle (`""_"-1"^0"e"`) is formed in the reaction. Therefore it’s a `β` negative decay.

In a `β` negative decay neutron turns into proton. Therefore the resultant atom should have

`"Protons = 23 + 1 = 24"`
`"Neutrons = (52 – 23) – 1 = 28"`

An atom with `24` protons is `"Cr"`

So, `β^-` decay of `""_23^52"V"` is

`""_23^52"V" -> ""_24^52"Cr" + ""_"-1"^0"e" + bar ν`

Note: In a `β`-decay (both positive and negative) mass number remains same.