In the nuclear reaction : `X (n, alpha)_3 Li^7` the term `X` will be `3`

Options

A. `""_5B^9`
B. `""_5B^10`
C. `""_5B^11`
D. `""_2He^4`

Answer

B

Explanation

In usual notation , the nuclear reaction X `(n, alpha) to ""_3Li^7` is written as
`""_ZX^A + ""_0n^1 to ""_2He^4 + ""_3Li^7 (alpha " particle is emitted ")`
Equation the values of Z and A on the sides , we get (conservation of Z and A )
Z +0 = 2 + 3 `therefore Z = 5`
and A +1 = 4 + 7 `therefore A = 10`
Thus , Z = 5 and A = 10 corresponding to Boron `(""_5B^10)`