Question

. When a uranium isotope \(_{9}^{235} 2 U\) is bombarded with a neutron, it generates \({ }_{36}^{89} K r\), three neutrons and :
(A) \({ }_{56}^{144} \mathrm{Ba}\)
B \({ }_{40}^{91} Z r\)
C \({ }_{36}^{101} K r\)
D \({ }_{36}^{103} K r\)

Answer 1

Solution:
$$
\begin{aligned}
&{ }_{92}^{235} U+{ }_{0}^{1} n \rightarrow{ }_{36}^{89} K r+3{ }_{0}^{1} n+X_{X}^{A} \\
&92+0=36+Z \\
&\Rightarrow Z=56 \\
&235+1=89+3+A \\
&\Rightarrow A=144
\end{aligned}
$$
So, \({ }_{56}^{144} \mathrm{Ba}\) is generated