Problem 47
Question
\({ }_{7} \mathrm{~N}^{13}\) changes to \({ }_{6} \mathrm{C}^{13}\) by emission of (a) electron (b) neutron (c) positron (d) proton
Step-by-Step Solution
Verified Answer
The emission is (c) positron.
1Step 1: Understand the Initial and Final Nuclei
We start with an atom \({ }_{7} \mathrm{~N}^{13}\), which means it has 7 protons and a mass number of 13. It changes to \({ }_{6} \mathrm{C}^{13}\), indicating it now has 6 protons but the same mass number of 13.
2Step 2: Compare Proton Numbers
The initial nucleus has 7 protons, while the final nucleus has 6 protons. This indicates a change where the number of protons reduces by 1.
3Step 3: Determine the Type of Emission
A decrease in the number of protons (from 7 to 6) while maintaining the same mass number (13) suggests an emission that transforms a proton into a neutron. This is characteristic of positron emission, where a proton turns into a neutron, emitting a positron.
Key Concepts
Positron EmissionProton to Neutron TransformationMass Number Conservation
Positron Emission
Positron emission is a type of nuclear decay where an unstable nucleus transforms by converting one of its protons into a neutron.
This process is accompanied by the emission of a positron, which is the antimatter counterpart of the electron and carries a positive charge.
Positron emission is a common way for isotopes with too many protons to stabilize. The nuclear equation for positron emission can be represented as:
This process is accompanied by the emission of a positron, which is the antimatter counterpart of the electron and carries a positive charge.
Positron emission is a common way for isotopes with too many protons to stabilize. The nuclear equation for positron emission can be represented as:
- A proton in the nucleus is converted into a neutron.
- A positron (β+) and a neutrino are emitted.
- The nitrogen-13 ( { }_{7} ext{N}^{13}) nucleus emits a positron and becomes carbon-13 ( { }_{6} ext{C}^{13}).
- The number of protons decreases by one, transforming nitrogen into carbon.
Proton to Neutron Transformation
The transformation of a proton into a neutron is a crucial part of certain types of radioactive decay, such as positron emission.
This process ensures that the nucleus becomes more energetically stable, especially when there is an excess of protons. In the proton to neutron transformation:
This process ensures that the nucleus becomes more energetically stable, especially when there is an excess of protons. In the proton to neutron transformation:
- One proton is lost from the nucleus, while one neutron is gained, maintaining a balance.
- This change does not affect the atomic mass but changes the atomic number.
- The proton number drops from 7 in nitrogen (resulting in carbon with 6 protons).
- There is no change in the mass number, which remains at 13, indicating stability in relative isotopic mass.
Mass Number Conservation
The principle of mass number conservation states that the total mass number before and after a nuclear reaction must remain the same.
This is a key concept in nuclear reactions, ensuring that the sum of protons and neutrons (mass number) in the reactants equals that in the products. During nuclear decay, such as positron emission:
This is a key concept in nuclear reactions, ensuring that the sum of protons and neutrons (mass number) in the reactants equals that in the products. During nuclear decay, such as positron emission:
- The mass number doesn't change even as an element transforms.
- A proton is converted to a neutron, hence the overall mass number (sum of protons and neutrons) is conserved.
- Nitrogen-13 ( { }_{7} ext{N}^{13}) changes to carbon-13 ( { }_{6} ext{C}^{13}), yet the mass number remains constant at 13.
- This showcases how nuclear transformations honor the mass number, despite changes in the atomic structure.
Other exercises in this chapter
Problem 45
\({ }_{2} \mathrm{U}^{238}\) is radioactive and it emits \(\alpha\) and \(\beta\) particles to form \({ }_{82} \mathrm{~Pb}^{206} .\) The number of \(\alpha\) a
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An element \(\mathrm{X}\) loses one \(\alpha\) and two two \(\beta\) particles in threc succossive stagcs. The resulting clements will be (a) an isotope of \(\m
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If uranium (mass number 238 and atomic number 92 ) emits an \(\alpha\) particle, the product has mass number and atomic number (a) 234 and 90 (b) 236 and 92 (c)
View solution Problem 49
Which of the following notations shows the product incorrectly? (a) \({ }_{5} \mathrm{~B}^{10}(\alpha, \mathrm{n}){ }_{7} \mathrm{~N}^{13}\) (b) \({ }_{96} \mat
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