Problem 57 - Chemistry: An Atoms-Focused Approach

Question

What are the products of radioactive decay of \(^{137} \mathrm{Cs} ?\) Write a balanced equation for the nuclear decay reaction.

Step-by-Step Solution

Answer

Answer: The radioactive decay of Cesium-137 (Cs-137) results in the creation of Barium-137 (Ba-137), a beta particle (electron), and an electron antineutrino. The balanced nuclear decay equation for this process would be: $$^{137}\mathrm{Cs} \rightarrow ^{137}\mathrm{Ba} + e^- + \bar{\nu}_e$$

Step 1: Identify the type of radioactive decay

In this case, \(^{137}\mathrm{Cs}\) undergoes beta decay, which means that a neutron within the nucleus will be converted into a proton, releasing an electron (also called a beta particle) and an electron antineutrino.

Step 2: Determine the products of the decay

As a result of beta decay, we will have one more proton and one fewer neutron in the nucleus. Since atomic number (number of protons) determines the element, we can predict that in the case of \(^{137}\mathrm{Cs}\), the resulting element will be Barium (\(\mathrm{Ba}\)) with an atomic number one higher than Cesium. The new nucleus will consist of 56 protons and 81 neutrons (137-56). The decay also produces a beta particle (electron) and an electron antineutrino, which are represented by \(e^-\) and \(\bar{\nu}_e\), respectively.

Step 3: Write the balanced nuclear decay equation

To write the balanced nuclear decay equation, we should make sure that the sum of the atomic masses and atomic numbers on both sides of the equation are equal. Atomic masses: \(137 (\mathrm{Cs}) = 137 (\mathrm{Ba})\) Atomic numbers: \(55 (\mathrm{Cs}) = 56 (\mathrm{Ba}) -1 (e^-)\) Considering these balances, we can now write the balanced nuclear decay equation for the radioactive decay of \(^{137}\mathrm{Cs}\): $$^{137}\mathrm{Cs} \rightarrow ^{137}\mathrm{Ba} + e^- + \bar{\nu}_e$$