Question

What is the particle $X$ in the following nuclear reaction :
$^{9}_{4}Be +^{4}_{2}He \to \,{}^{12}_{6} C +X$

• A. electron
• B. proton
• C. Photon
• D. Neutron

Answer 1

Answer: (D)

$^{9}_{4}Be +^{4}_{2}He \to ^{12}_{6} C +^{1}_{o}X$
Comparing sum of mass numbers and atomic numbers on both sides, we get
$A = 1, Z = 0$
Hence $X$ represents neutron $\, \left(^{1}_{o}n\right)$ (neutron).