A) \[4\alpha ,1\beta \]
B) \[4\alpha ,2\beta \]
C) \[5\alpha ,1\beta \]
D) \[5\alpha ,2\beta \]
Correct Answer: A
Solution :
\[_{90}T{{h}^{228}}{{\xrightarrow[{}]{{}}}_{83}}B{{i}^{212}}{{+}_{{{x}_{2}}}}H{{e}^{4}}+{{y}_{1}}{{e}^{0}}\] On comparison mass number \[228=212+4x\] \[x=\frac{228-212}{4}=4\] On comparison atomic number \[90=83+2x-y\] \[90=83+8-y\] \[y=91-90\] \[y=1\] Hence, number of\[\alpha -\]particle = 4 Number of\[\beta -\]particle = 1
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