A) 3
B) 4
C) 6
D) 2
Correct Answer: A
Solution :
Number of \[\alpha \text{-}\]particles\[=\frac{change\text{ }in\text{ }atomic\text{ }mass}{4}\]\[=\frac{218-206}{4}=\frac{12}{4}=3\]You need to login to perform this action.
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