When a `._4 Be^9` atom is bombarded with `prop-`particle, one of the product of nuclear transmutation is `._6C^12`. The other is.

Options

A. `""_1D^2`
B. `""_0n^1`
C. `""_(-1)e^0`
D. `""_1H^1`

Answer

B

Explanation

`""_4Be^9 + ""_2He^4 to ""_6C^12 + X`
For conservation of charge, 4+2=6+0
`therefore` For X,Z =0
For conservation of mass , 9 + 4 =12 +1
`therefore` For X,A = 1
`therefore` The particle is `""_0n^1` (neutron) .