The beta decay of Francium-223 (Fr-223) involves the transformation of a neutron into a proton, with the emission of a beta particle (electron) and an antineutrino. The resulting product is Radium-223 (Ra-223).
The nuclear equation for the beta decay of Francium-223 (Fr-223) can be represented as follows:
Breaking down the equation:
The initial nuclide is Francium-223 with 87 protons and 223 nucleons.
In the decay process, a neutron within the nucleus is transformed into a proton.
The emission of a beta particle (
) occurs, which is essentially an electron (
)
An antineutrino (νˉe ) is also emitted to conserve energy and momentum.
The resulting nuclide is Radium-223 (Ra-223) with 88 protons and 223 nucleons.
This beta decay process is driven by the conversion of a neutron into a proton, leading to an increase in atomic number by one unit. The emitted beta particle carries away the excess negative charge, and the antineutrino is released to conserve energy during the transformation. Overall, this nuclear equation represents the beta decay of Francium-223 into Radium-223.
The question probable may be:
How do you write the nuclear equation for the beta decay of francium-223 87fr?