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 \[\alpha \] particles lost \[=\frac{\begin{align} & atomic\,\,mass\,\,of\,\,reactant-atomic \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mass\,\,of\,\,product \\ \end{align}}{4}\] (ii) Number of \[\beta -\]particles \[=(2\times number\,\,of\,\,\alpha -particles)-(differnce\,\,in\]\[atomic\,\,number\,\,of\,\,product\,\,and\,\,reactant)\] 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 \[6\beta \] particles.
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