Question

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uestion 2) neutron bombardment of uranium
an induce the reaction below. nuclide x is
hich of the following?
$$ _ { 92 } ^ { 235 } mathrm { u } + _ { 0 } ^ { 1 } n longrightarrow _ { 55 } ^ { 141 } mathrm { cs } + 3 _ { 0 } ^ { 1 } n + x $$

Explanation:

Step1: Calculate the mass number of \(X\)

According to the conservation of mass number (sum of mass numbers on the left - hand side = sum of mass numbers on the right - hand side).
The mass number of \(^{235}_{92}U\) is \(A_{U}=235\), the mass number of \(^{1}_{0}n\) is \(A_{n}=1\), the mass number of \(^{141}_{55}Cs\) is \(A_{Cs} = 141\), and the mass number of \(3\) neutrons (\(3^{1}_{0}n\)) is \(3\times1=3\).
Let the mass number of \(X\) be \(A_{X}\). Then \(235 + 1=141+3 + A_{X}\), so \(A_{X}=236-(141 + 3)=92\).

Step2: Calculate the atomic number of \(X\)

According to the conservation of atomic number (sum of atomic numbers on the left - hand side = sum of atomic numbers on the right - hand side).
The atomic number of \(^{235}_{92}U\) is \(Z_{U}=92\), the atomic number of \(^{1}_{0}n\) is \(Z_{n}=0\), the atomic number of \(^{141}_{55}Cs\) is \(Z_{Cs}=55\), and the atomic number of \(3\) neutrons (\(3^{1}_{0}n\)) is \(3\times0 = 0\).
Let the atomic number of \(X\) be \(Z_{X}\). Then \(92+0=55 + 0+Z_{X}\), so \(Z_{X}=92 - 55=37\).

Step3: Identify the element

The element with atomic number \(Z = 37\) is rubidium (\(Rb\)). So \(X\) is \(^{92}_{37}Rb\).

Answer:

\(^{92}_{37}Rb\)