To solve the problem of determining the balanced equation for the beta minus emission of phosphorus-32, let's understand what beta minus emission entails and analyze each given option to find the correct answer.
### Understanding Beta Minus Emission:
Beta minus emission, often referred to as beta decay, is a type of radioactive decay in which a neutron in an atom's nucleus converts into a proton while emitting an electron (beta particle) and an antineutrino.
Key points for beta minus emission:
1. The atomic number (number of protons) of the atom increases by 1, because a neutron is converted into a proton.
2. The mass number (total number of protons and neutrons) remains unchanged.
3. An electron ([tex]${ }_{-1}^0 e$[/tex]) is emitted from the nucleus.
For phosphorus-32, with an initial atomic number of 15 and mass number of 32, the decay can be written as:
[tex]\[ { }_{15}^{32} P \longrightarrow \text{New Element} + { }_{-1}^0 e \][/tex]
### Analyzing Each Option:
1. [tex]\[ { }_{15}^{32} P \longrightarrow{ }_{13}^{28} Al +{ }_2^4 He \][/tex]
- Aluminum-28 and Helium-4 are produced.
- The atomic number decreases from 15 to 13, indicating alpha decay, not beta minus emission. Incorrect.
2. [tex]\[ { }_{15}^{32} P \longrightarrow{ }_{14}^{32} Si +{ }_{-1}^0 e \][/tex]
- Silicon-32 is produced.
- The atomic number increases from 15 to 14 and an electron is emitted. This fits the beta minus decay process where the atomic number increase should be from 15 to 16. Incorrect.
3. [tex]\[ { }_{15}^{32} P \longrightarrow{ }_{16}^{32} S +{ }_{-1}^0 e \][/tex]
- Sulfur-32 is produced.
- The atomic number increases from 15 to 16 and an electron is emitted. This is a perfect fit for beta minus emission. Correct.
4. [tex]\[ { }_{15}^{32} P \longrightarrow{ }_{14}^{32} Si +{ }_{+1}^0 e \][/tex]
- Silicon-32 is produced with a positron.
- The atomic number increases from 15 to 14 and a positron ([tex]${ }_{+1}^0 e$[/tex]) is emitted, indicating beta plus (positron) emission, not beta minus emission. Incorrect.
### Conclusion:
The balanced equation that represents the beta minus emission of phosphorus-32 is:
[tex]\[ { }_{15}^{32} P \longrightarrow{ }_{16}^{32} S +{ }_{-1}^0 e \][/tex]
Thus, the correct answer is:
[tex]\[ {3} \][/tex]
