High School

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A radioactive nucleus undergoes beta plus decay. The remaining daughter nucleus is [tex]$Au-193$[/tex].

What was the identity of the parent nucleus?

Choose one answer:

A. [tex]$Pt-193$[/tex]

B. [tex]$Hg-193$[/tex]

C. [tex]$Tl-197$[/tex]

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).