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The Balanced Nuclear Equation for Alpha Decay Of 218 Po { } ^ { 218 } \text { Po }

Question 32

Multiple Choice

The balanced nuclear equation for alpha decay of 218 Po { } ^ { 218 } \text { Po } is _____.


A) 84218Po85218At+10β{ } _ { 84 } ^ { 218 } \mathrm { Po } \rightarrow { } _ { 85 } ^ { 218 } \mathrm { At } + { } _ { - 1 } ^ { 0 } \beta
B) 84218Po80212Hg+42α{ } _ { 84 } ^ { 218 } \mathrm { Po } \rightarrow { } _ { 80 } ^ { 212 } \mathrm { Hg } + { } _ { 4 } ^ { 2 } \alpha
C) 84218Po82214 Pb+24α{ } _ { 84 } ^ { 218 } \mathrm { Po } \rightarrow { } _ { 82 } ^ { 214 } \mathrm {~Pb} + { } _ { 2 } ^ { 4 } \alpha
D) 84218Po83218Bi++10β{ } _ { 84 } ^ { 218 } \mathrm { Po } \rightarrow { } _ { 83 } ^ { 218 } \mathrm { Bi } + { } _ { + 1 } ^ { 0 } \beta
E) 84218Po84214Bi+24α{ } _ { 84 } ^ { 218 } \mathrm { Po } \rightarrow { } _ { 84 } ^ { 214 } \mathrm { Bi } + { } _ { 2 } ^ { 4 } \alpha

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