In the nuclear reaction \( {}^{4}_{2}\mathrm{He}+{}^{A}_{Z}\mathrm{X}\rightarrow{}^{A+3}_{Z+2}\mathrm{Y}+\mathrm{b} \), the particle \(\mathrm{b}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Neutron
In a nuclear reaction, the total number of nucleons and the total charge stay the same on the two sides. So the mass numbers must balance, and the atomic numbers must balance too. Let the particle be \( {}^{A'}_{Z'}\mathrm{b} \).
Balancing the mass numbers,
\( \begin{aligned} 4+A &= (A+3)+A' \\ A' &= 1 \end{aligned} \)
Balancing the atomic numbers,
\( \begin{aligned} 2+Z &= (Z+2)+Z' \\ Z' &= 0 \end{aligned} \)
From the above, the particle has mass number 1 and no charge, \( {}^{1}_{0}\mathrm{n} \). Hence, b is a neutron.
NCERT (2026-27):
Example 13.4 – “… However, the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. …”