A) \[7\alpha ,10\beta \]
B) \[8\alpha ,6\beta \]
C) \[5\alpha ,6\beta \]
D) \[6\alpha ,8\beta \]
Correct Answer: B
Solution :
\[\underset{92}{\mathop{238}}\,U\to \underset{82}{\mathop{206}}\,Pb\] |
Number of \[\alpha \]-particles |
\[=\frac{chnages\,\,in\,\,mass\,\,number}{4}\] |
\[=\frac{238-206}{4}\] |
\[=\frac{38}{4}=8\alpha \] |
Number of \[\beta \]-particles \[=2\times \] number of \[\alpha \]-particles - change in atomic number |
\[=2\times 8-\left( 92-82 \right)\] |
\[=16-10=6\beta \] |
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