A) \[6\] and \[8\]
B) \[8\] and \[10\]
C) \[5\] and \[8\]
D) \[8\] and \[6\]
Correct Answer: D
Solution :
\[_{Y}{{A}^{X}}\xrightarrow[\beta =n]{\alpha =m}\,_{Y-10}^{X-32}B\] given, \[\alpha -particle=m,\]\[\beta -particle=n\] difference between \[m(\alpha -particle)=\frac{at.\,\,wt.\,\,of\,\,A\,and\,B}{4}\] \[=\frac{32}{4}=8\alpha particles\] \[n(\beta -particle)=2\times \alpha -particle-difference\] between at. no. of A and B \[=2\times 8-10\] \[=16-10\] \[=6\beta -particles\] So, the value of \[m=8,\] the value of \[n=6.\].You need to login to perform this action.
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