In the following nuclear reaction,
\( {}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{a}_{b}\mathrm{Sr} + 2\,{}^{1}_{0}\mathrm{n} \) we have
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( a = 94,\ b = 38 \)
In a nuclear reaction the mass number (the sum of the numbers of nucleons) and the atomic number (the charge) are both conserved. Here \( a \) is the mass number and \( b \) is the atomic number of strontium.
Mass number:
\( \displaystyle 1 + 235 = 140 + a + 2(1) \;\Rightarrow\; 236 = 142 + a \;\Rightarrow\; a = 94 \)
Atomic number:
\( \displaystyle 0 + 92 = 54 + b + 2(0) \;\Rightarrow\; b = 38 \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Fission – “… A most important neutron-induced nuclear reaction is fission.”
Fission – “… The same reaction can produce other pairs of intermediate mass fragments
\( \displaystyle {}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{236}_{92}\mathrm{U} \rightarrow {}^{133}_{51}\mathrm{Sb} + {}^{99}_{41}\mathrm{Nb} + 4\,{}^{1}_{0}\mathrm{n} \qquad (13.11) \)
Or, as another example,
\( \displaystyle {}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2\,{}^{1}_{0}\mathrm{n} \qquad (13.12) \)”