Answer :
To identify the balanced nuclear equation for the beta plus decay ([tex]\({ \beta^+ }\)[/tex] decay or positron emission) of [tex]\( \mathrm{C}-11 \)[/tex], let's start by understanding the process of beta plus decay. During beta plus decay:
1. A proton in the nucleus is converted into a neutron.
2. This process emits a positron ([tex]\({ }_{+1}^0 e\)[/tex]) and a neutrino (which is often not shown in the equation).
The key points to remember are:
- The atomic number ([tex]\(Z\)[/tex]) of the original atom decreases by 1.
- The mass number ([tex]\(A\)[/tex]) remains unchanged because the mass doesn't significantly change when a proton converts to a neutron.
Given these principles, let's analyze the given options:
1. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{+1}^0 e\)[/tex]
- Here, the mass number (11) remains unchanged.
- The atomic number decreases from 6 (Carbon) to 5 (Boron).
- Emission of a positron ([tex]\({ }_{+1}^0 e\)[/tex]) is shown.
2. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_7^{11} \mathrm{N} +{ }_{+1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number increases from 6 to 7, which is not correct for beta plus decay.
3. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_7^{11} \mathrm{N} +{ }_{-1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number increases from 6 to 7, and it shows the emission of an electron ([tex]\({ }_{-1}^0 e\)[/tex]), which represents beta minus decay, not beta plus decay.
4. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{-1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number decreases from 6 to 5, but it shows the emission of an electron ([tex]\({ }_{-1}^0 e\)[/tex]), which is incorrect for beta plus decay.
Among the given options, the correct nuclear equation for beta plus decay of [tex]\({ }_6^{11} \mathrm{C}\)[/tex] is:
[tex]\[ { }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{+1}^0 e \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1} \][/tex]
1. A proton in the nucleus is converted into a neutron.
2. This process emits a positron ([tex]\({ }_{+1}^0 e\)[/tex]) and a neutrino (which is often not shown in the equation).
The key points to remember are:
- The atomic number ([tex]\(Z\)[/tex]) of the original atom decreases by 1.
- The mass number ([tex]\(A\)[/tex]) remains unchanged because the mass doesn't significantly change when a proton converts to a neutron.
Given these principles, let's analyze the given options:
1. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{+1}^0 e\)[/tex]
- Here, the mass number (11) remains unchanged.
- The atomic number decreases from 6 (Carbon) to 5 (Boron).
- Emission of a positron ([tex]\({ }_{+1}^0 e\)[/tex]) is shown.
2. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_7^{11} \mathrm{N} +{ }_{+1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number increases from 6 to 7, which is not correct for beta plus decay.
3. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_7^{11} \mathrm{N} +{ }_{-1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number increases from 6 to 7, and it shows the emission of an electron ([tex]\({ }_{-1}^0 e\)[/tex]), which represents beta minus decay, not beta plus decay.
4. [tex]\({ }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{-1}^0 e\)[/tex]
- The mass number remains unchanged.
- The atomic number decreases from 6 to 5, but it shows the emission of an electron ([tex]\({ }_{-1}^0 e\)[/tex]), which is incorrect for beta plus decay.
Among the given options, the correct nuclear equation for beta plus decay of [tex]\({ }_6^{11} \mathrm{C}\)[/tex] is:
[tex]\[ { }_6^{11} \mathrm{C} \longrightarrow{ }_5^{11} \mathrm{B} +{ }_{+1}^0 e \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1} \][/tex]
