Question

complete the following nuclear equation.
\ce{_{94}^{239}pu -> \square + _{2}^{4}he}

Explanation:

Step1: Balance mass number

The mass number of \(_{94}^{239}\text{Pu}\) is \(239\), and the mass number of \(_{2}^{4}\text{He}\) is \(4\). Let the mass number of the unknown nuclide be \(A\). Then \(239=A + 4\), so \(A=239 - 4=235\).

Step2: Balance atomic number

The atomic number of \(_{94}^{239}\text{Pu}\) is \(94\), and the atomic number of \(_{2}^{4}\text{He}\) is \(2\). Let the atomic number of the unknown nuclide be \(Z\). Then \(94=Z + 2\), so \(Z=94 - 2 = 92\). The element with atomic number \(92\) is uranium (\(U\)).

Answer:

\(_{92}^{235}\text{U}\)