Answer :
To determine the identity of the parent nucleus that underwent beta plus decay, let's take a closer look at what happens during this type of decay.
Beta Plus Decay:
1. In beta plus decay, a proton in the nucleus is converted into a neutron, and a positron is emitted.
2. As a result, the atomic number (the number of protons) of the nucleus decreases by 1.
3. The mass number (the total number of protons and neutrons) stays the same.
Given Information:
- The daughter nucleus after the decay is [tex]\( \text{Au-193} \)[/tex].
Gold (Au) Information:
- Gold (Au) has an atomic number of 79.
Understanding the Decay:
- Since the atomic number decreases by 1 during beta plus decay, the parent nucleus must have had an atomic number of 80. The mass number remains 193.
Using the Periodic Table:
- We need to find which element has an atomic number of 80.
- Mercury (Hg) has an atomic number of 80.
Parent Nucleus Identity:
- Since the atomic number of the parent nucleus is 80 and the mass number is 193, the parent nucleus must be [tex]\( \text{Hg-193} \)[/tex].
Thus, the correct identity of the parent nucleus is [tex]\( \text{Hg-193} \)[/tex], matching option (B).
Beta Plus Decay:
1. In beta plus decay, a proton in the nucleus is converted into a neutron, and a positron is emitted.
2. As a result, the atomic number (the number of protons) of the nucleus decreases by 1.
3. The mass number (the total number of protons and neutrons) stays the same.
Given Information:
- The daughter nucleus after the decay is [tex]\( \text{Au-193} \)[/tex].
Gold (Au) Information:
- Gold (Au) has an atomic number of 79.
Understanding the Decay:
- Since the atomic number decreases by 1 during beta plus decay, the parent nucleus must have had an atomic number of 80. The mass number remains 193.
Using the Periodic Table:
- We need to find which element has an atomic number of 80.
- Mercury (Hg) has an atomic number of 80.
Parent Nucleus Identity:
- Since the atomic number of the parent nucleus is 80 and the mass number is 193, the parent nucleus must be [tex]\( \text{Hg-193} \)[/tex].
Thus, the correct identity of the parent nucleus is [tex]\( \text{Hg-193} \)[/tex], matching option (B).
