A) \[4\alpha \]and\[1\beta \]
B) \[3\alpha \]and\[7\beta \]
C) \[8\alpha \]and\[1\beta \]
D) \[4\alpha \]and\[7\beta \]
Correct Answer: A
Solution :
Number of\[\alpha -\]particles \[=\frac{change\text{ }in\text{ }mass\text{ }number}{4}\] \[=\frac{228-212}{4}\] \[=\frac{16}{4}=4\] Number of\[\beta \]particles \[=2\alpha -\] (change in atomic number) \[=2\times 4-(90-83)\] \[=8-7=1\] \[\therefore \]no. of \[\alpha -\]particles emitted = 4 and\[\beta \]particles = 1You need to login to perform this action.
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