Question

complete this equation that represents the process of nuclear fusion.
_{88}^{226}ra \to _{b}^{a}rn + _{2}^{4}he
a:
b:

Explanation:

Step1: Calculate the mass number (A)

In a nuclear reaction, the sum of mass numbers on the left - hand side equals the sum on the right - hand side.
The mass number of \(^{226}_{88}Ra\) is \(226\), and the mass number of \(^{4}_{2}He\) is \(4\). Let the mass number of \(Rn\) be \(A\). Then \(226=A + 4\).

$$A=226 - 4=222$$

Step2: Calculate the atomic number (B)

In a nuclear reaction, the sum of atomic numbers on the left - hand side equals the sum on the right - hand side.
The atomic number of \(^{226}_{88}Ra\) is \(88\), and the atomic number of \(^{4}_{2}He\) is \(2\). Let the atomic number of \(Rn\) be \(B\). Then \(88=B + 2\).

$$B=88 - 2=86$$

Answer:

A: \(222\)
B: \(86\)