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

The Correct Option is A

The given nuclear reaction involves the fission of a uranium isotope \(^{235}_{92}U\) when it is bombarded with a neutron \(^{1}_{0}n\). During this process, it is converted into \(^{89}_{36}Kr\), three neutrons (3 \(^{1}_{0}n\)), and another nucleus. Let's solve the problem step-by-step:

1. Initially, we have uranium-235 and a neutron involved in the reaction:

\(^{235}_{92}U + ^{1}_{0}n \rightarrow \text{Products}\)

1. Solving this, the reaction gives krypton-89, three neutrons, and an unknown nucleus:

\(^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr + ^{144}_{56}Ba + 3\ {}^{1}_{0}n\)

1. We need to balance this nuclear equation both in terms of mass numbers (A) and atomic numbers (Z):
2. Analyze Mass Number Balance:

\(235 + 1 = 89 + (X) + 3 \times 1\)

\(236 = 89 + (X) + 3\)

Solving for X, the unknown mass number:

\(X = 236 - 89 - 3 = 144\)

1. Analyze Atomic Number Balance:

\(92 = 36 + (Z) + 3 \times 0\)

Solving for Z, the unknown atomic number:

\(Z = 92 - 36 = 56\)

1. From the balancing of both mass and atomic numbers, we identify that the unknown product is \(^{144}_{56}Ba\). Thus, \(^{144}_{56}Ba\) is the correct missing nucleus.

Therefore, the balanced nuclear fission reaction is:

\(^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr + ^{144}_{56}Ba + 3\ {}^{1}_{0}n\)

The correct answer is indeed \(^{144}_{56}Ba\).