Question

When a uranium isotope $^{235}_92 U$ is bombarded with a neutron, it generates $^{89}_{36} Kr$, three neutrons and :

• A. $^{144}_{56} Ba$
• B. $^{91}_{40} Zr$
• C. $^{101}_{36} Kr$
• D. $^{103}_{36} Kr$

Answer 1

Answer: (A)

$^{144}_{56} Ba$

Answer 2

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr+3 ^{1}_{0}n+X_{X}^{A}$

$92+0=36+Z$

$\Rightarrow Z=56$

$235+1=89+3+A$

$\Rightarrow A=144$

So, $^{144}_{56}Ba$ is generated