Question

How do you write the nuclear equation for the beta decay of francium-223?

Answer 1

Hint:To solve this we must know the atomic number of francium-223. We must also know the reaction for beta-decay. When a radioactive nucleus undergoes beta-decay, the mass number of the element remains the same but the number of protons increases by 1.

Complete answer:
We know that the general representation of elements is ${}_{\text{Z}}^{\text{A}}{\text{X}}$. Where ${\text{Z}}$ is the atomic number of the element, ${\text{A}}$ is the mass number of the element and ${\text{X}}$ is the atomic symbol of the element.
We know that the atomic number is equal to the number of protons or to the number of electrons.
When a radioactive nucleus undergoes beta-decay, the mass number of the element remains the same but the number of protons increases by 1 i.e. the atomic number increases by 1.
The atomic number of francium is 87 and the mass number of francium is 223. Thus, the element francium is represented as ${}_{87}^{233}{\text{Fr}}$.
When francium i.e. ${}_{87}^{233}{\text{Fr}}$ undergoes beta-decay its mass number remains the same but the atomic number increases by 1. Thus, a new element with atomic number 88 is formed. The element having atomic number 88 is radium. The chemical symbol for radium is ${\text{Ra}}$.
Thus, the nuclear equation for the beta decay of francium-223 is as follows:
${}_{87}^{233}{\text{Fr}} \to {}_{88}^{233}{\text{Ra}} + {}_{ - 1}^0e + \nu $
During beta-decay an electron and a neutrino are emitted. The emitted electron helps in conserving the charge.

Note: The general reaction for beta-decay is ${}_{\text{Z}}^{\text{A}}{\text{X}}\xrightarrow{{{\beta }}}{}_{{\text{Z}} + 1}^{\text{A}}{\text{Y}}$. Here, A is the atomic mass and Z is the atomic number. During beta-decay an electron is emitted. This emitted electron helps in conserving the charge. Along with an electron, neutrino is also emitted.