A) \[7\alpha ,5\beta \]
B) \[6\alpha ,4\beta \]
C) \[4\alpha ,3\beta \]
D) \[8\alpha ,6\beta \]
Correct Answer: D
Solution :
Key Idea: (i) Number of a particles lost \[\begin{align} & =\underline{\begin{align} & \text{atomic}\,\text{mass}\,\text{of}\,\text{reatant}-\text{atomic} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{mass of}\,\text{product} \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \\ \end{align}\] (ii) Number of \[\,\beta -\]particles \[=(2\,\times \,\text{number}\,\text{of}\,\alpha -particles)-\] \[(difference\,in\,atomic\,number\,of\,product\,and\,\text{reactant})\]Given \[{{\,}_{92}}{{U}^{238}}\xrightarrow{{}}{{\,}_{82}}P{{b}^{206}}\] \[\therefore \]Number of\[\alpha -\]particles \[=\frac{238-206}{4}=\frac{32}{4}=8\] \[\therefore \]Number of \[\beta -\]particles \[=(2\,\times \,8)-(92-82)\] \[=16-10=6\] \[\therefore \]Total of \[8\alpha \]and \[6\alpha \]particles.
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