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In the following reaction. .12 Mg^24 +...

In the following reaction.
`._12 Mg^24 + ._2He^4 rarr ._14 S i^X +._0 n^1, X` is.

Text Solution

Verified by Experts

The correct Answer is:
B
`""_(12)^(24)Mg+""_(2)^(4)He rarr ""_(14)^(x)Si+""_(0)^(1)n`
According to mass number conservation, we get
`24+4=x+1`
`or x=27`
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