Question

complete this equation that represents the process of nuclear fission.
_{94}^{239}pu+_{0}^{1}n\to_{b}^{a}ba+_{38}^{91}sr+3_{0}^{1}n
a:
b:

Explanation:

Step1: Calculate the mass number (A)

In nuclear reactions, the sum of mass numbers on the left - hand side equals the sum of mass numbers on the right - hand side.
The mass number of \(_{94}^{239}Pu\) is \(239\), and the mass number of \(_{0}^{1}n\) is \(1\). On the right - hand side, the mass number of \(_{38}^{91}Sr\) is \(91\), and the mass number of \(3\) \(_{0}^{1}n\) is \(3\times1 = 3\).
Let the mass number of \(Ba\) be \(A\). Then \(239+1=A + 91+3\).

$$A=239 + 1-(91 + 3)=146$$

Step2: Calculate the atomic number (B)

In nuclear reactions, the sum of atomic numbers on the left - hand side equals the sum of atomic numbers on the right - hand side.
The atomic number of \(_{94}^{239}Pu\) is \(94\), and the atomic number of \(_{0}^{1}n\) is \(0\). On the right - hand side, the atomic number of \(_{38}^{91}Sr\) is \(38\), and the atomic number of \(_{0}^{1}n\) is \(0\).
Let the atomic number of \(Ba\) be \(B\). Then \(94+0=B + 38+0\).

$$B=94 - 38=56$$

Answer:

A: \(146\)
B: \(56\)