A) \[\frac{x-m}{4}\]
B) \[n+\frac{x-m}{2}+y\]
C) \[n+\frac{x-m}{2}-y\]
D) \[2y-n+x-m\]
Correct Answer: C
Solution :
For the transformation \[_{y}^{x}A\] to \[_{n}^{m}B\], the complete nuclear reaction is \[_{y}^{x}A\xrightarrow{{}}_{n}^{m}B+{{p}_{2}}H{{e}^{4}}+{{q}_{-1}}{{\beta }^{o}}\] [Let p and q are the number of \[\alpha \] and \[\beta \]particles emitted respectively.] On comparing atomic mass, we get \[x=m+4p+0\] \[p=\frac{x-m}{4}\] On comparing atomic number, we get \[y=n+2p-q\] \[y=n+\frac{2(x-m)}{4}-q\] \[=n=\frac{x-m}{2}-q\] \[\therefore \] \[q=n+\frac{x-m}{2}-y\]
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