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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.

A

28

B

22

C

27

D

26

Text Solution

Verified by Experts

The correct Answer is:
C
By using the conservation of mass number, for the reaction
`""_12Mg^24 + ""_2He^4 to ""_14 Si^x + ""_0n^1`
we get, 24 + 4 = x + 1 `therefore x =27`
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