When a deuterium is bombarded on `._(8)O^(16)` nucleus, an `alpha`-particle is emitted, then the product nucleus is

Options

A. `""_5B^10`
B. `""_7N^14`
C. `""_7^13`
D. `""_4Be^9`

Answer

B

Explanation

The nuclear reaction is `""_8O^16 + ""_1D^2 to ""_1He^4 + ""_ZX^A`
Using conservation of mass, we get
16 + 2 = 4 + A `therefore` A = 14
and for conservation of Z, 8+1=2+Z `therefore Z = 7`
`therefore` the unknown nucleus is `""_7N^14` .