Answer :
Final answer:
The balanced nuclear reaction for iodine-131 decaying to xenon-131 is written with iodine-131 on the left, followed by the formation of xenon-131, a beta particle, and a gamma ray on the right.
Explanation:
The balanced nuclear reaction for the decay of iodine-137 (which is an error, as the student likely meant iodine-131) to xenon-137 is as follows:
_{53}^{131}I \rightarrow _{54}^{131}Xe + _{-1}^{0}e (\beta^-) + \gamma
In this equation, iodine-131 undergoes beta decay to form xenon-137. When iodine-131 decays, it emits a beta particle (electron) and a gamma ray, and the resultant daughter isotope is xenon-131. Remember to ensure the atomic numbers and mass numbers are balanced on both sides of the reaction.
Explanation:
First off, it is important to know the type of decay that Iodine undergo to give Xenon.
This is achieved by comparing the mass number and atomic number of the products and reactants.
Iodine
Mass Number = 137
Atomic Number = 53
Xenon
Mass Number = 137
Atomic Number = 54
Upon comparison, we can tell that the only change is the increase in the atomic numbers.
Due to this, we now know it is a beta decay. In beta decay, one of the neutrons in the nucleus suddenly changes into a proton, causing an increase in the atomic number of an element.
In a balanced nuclear equation, the sum of the mass numbers on the reactant side must equal the sum of the mass numbers on the product side. The same must be true for the atomic numbers.
Reactant side;
Mass number = 137
Atomic Number = 53
Product side;
Mass number = 137 +0 = 137
Atomic Number = 54 + (-1) = 53
The equation is given in the attachment.
