Question

What is the particle $X$ in the following nuclear reaction :
${}_4^{9}Be{+}_2^{4}He\to {}_6^{12}C+X$

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

Answer 1

Answer: (D)

${}_4^{9}Be{+}_2^{4}He{\to }_6^{12}C{+}_o^{1}X$
Comparing sum of mass numbers and atomic numbers on both sides, we get
$A=1,Z=0$
Hence $X$ represents neutron $\left({}_o^{1}n\right)$ (neutron).