[tex]$\beta$[/tex] - decay producing ${ }_{56}^{137} Ba$
Answer (1)
The problem states that 56 137 B a is produced by β -decay.
In β − -decay, the atomic number increases by 1, while the mass number remains the same.
Therefore, the parent nuclide has a mass number of 137 and an atomic number of 56 − 1 = 55 .
The element with atomic number 55 is Cesium (Cs), so the parent nuclide is 55 137 C s .
Explanation
Understanding the Problem We are given that a β -decay produces 56 137 B a . This means that a parent nucleus undergoes β -decay and transforms into Barium-137, which has an atomic number of 56 and a mass number of 137. Our goal is to identify this parent nucleus.
The Process of Beta-Minus Decay In β − -decay, a neutron within the nucleus converts into a proton, emitting an electron and an antineutrino. This process increases the atomic number (number of protons) by 1, while the mass number (total number of protons and neutrons) remains the same.
Determining Atomic and Mass Numbers Let's denote the parent nuclide as Z A X , where A is the mass number and Z is the atomic number. After β − -decay, it transforms into 56 137 B a . This implies that the mass number A of the parent nuclide is 137, and its atomic number Z must be one less than that of Barium since the atomic number increases by 1 during the decay.
Identifying the Parent Nuclide Since the atomic number of Barium (Ba) is 56, the atomic number Z of the parent nuclide is 56 − 1 = 55 . The element with atomic number 55 is Cesium (Cs). Therefore, the parent nuclide is 55 137 C s .
Examples
Beta decay is crucial in nuclear medicine, particularly in diagnostic imaging and cancer therapy. For instance, radioactive isotopes like Cesium-137 ( 137 C s ) undergo beta decay, emitting particles that can be detected or used to target cancerous cells. Understanding beta decay helps medical professionals choose appropriate isotopes, calculate dosages, and predict the effectiveness of treatments, ensuring safer and more precise medical interventions.
