A) \[\frac{A-Z-4}{Z-2}\]
B) \[\frac{A-Z-8}{Z-4}\]
C) \[\frac{A-Z-4}{Z-8}\]
D) \[\frac{A-Z-12}{Z-4}\]
Correct Answer: C
Solution :
\[_{Z}^{A}X\xrightarrow[{}]{(3\alpha +2positron)}_{Z-3\times 2-2\times 1}^{A-3\times 4}\text{X}=_{Z-8}^{A-12}\text{X}\] \[\therefore \] \[\therefore \,\,\frac{No.\,of\,Neutrons}{No.\,of\,\Pr otons}\,=\frac{(A-12)-(Z-8)}{Z-8}\] \[=\frac{A-Z-4}{Z-8}\]
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