Question

$N^{14}$ is bombarded with ${}_{2}H{e}^4$. The resulting nucleus is ${}_8{O}^{17}$ with the emission of

• A. neutrino
• B. antineutrino
• C. proton
• D. neutron

Answer 1

Answer: (C)

Mass number and atomic number in nuclear reactions are conserved.
The nuclear reaction is as follows.
${}_7{N}^{14}+{}_{2}H{e}^4⟶{}_8{O}^{17}+{}_q{X}^P$
Conservation of mass number gives
$p=14+4-17=1$
Conservation of atomic number gives
$q=7+2-8=1$
Hence, particle is a proton ${}_1{H}^1$.