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

Answer:

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

Explanation:

Answer 2

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.