Question

Which of the following represents a balanced nuclear equation showing bismuth-212 undergoing alpha decay followed by beta decay? (2 points)

superscript 212 over subscript 83 Bi yields superscript 4 over subscript 2 He + superscript 208 over subscript 81 Tl yields x−superscript 208 over subscript 82 Pb + superscript 0 over subscript -1 Beta

superscript 212 over subscript 83 Bi + superscript 0 over subscript -1 Beta yields superscript 4 over subscript 2 He + superscript 208 over subscript 81 Tl yields x−superscript 208 over subscript 82 Pb

superscript 216 over subscript 85 As yields superscript 4 over subscript 2 He + superscript 216 over subscript 85 As yields superscript 216 over subscript 85 Pb + superscript 0 over subscript -1 Beta

superscript 212 over subscript 83 Bi yields superscript 0 over subscript -1 Beta + superscript 208 over subscript 81 Tl yields superscript 208 over subscript 82 Pb + superscript 4 over subscript 2 He

Answer 1

The reactions are

1)

`Bi^(212) _(83) --> He^(4) _(2) +Tl^(208) _(81) ---> Pb^(208) _(82)+\beta ^(0) _(-1)`

In this reaction the Bismuth gives an alpha particle and followed by beta decay

2)

`Bi^(212) _(83) + \beta ^(0)_(-1) -->He^(4) _(2)+Tl^(208) _(81)  --->Pb^(208) _(82) \\`

In this reaction the Bismuth is undergoing beta bombardment followed by alpha decay.

3)

`As^(216) _(85)--> He^(4) _(2)+As^(216) _(85) --->Pb^(216) _(85) +\beta ^(0)_(-1)`

This is not a reaction of Bismuth.

4)

`Bi^(212) _(83) --->\beta ^(0)_(-1) +Tl^(208) _(81)--->Pb^(208) _(82)+He^(4) _(2)`

This is first beta elimination followed by alpha elimination.