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Sagot :
To determine which particle needs to be added to the equation \( \, ^3_1 H \rightarrow \, ^3_2 He + \text{ ? } \) to ensure the total numbers of neutrons and protons remain unchanged, let's analyze the decay process in detail.
1. Determine the Composition of Tritium (\( \, ^3_1 H \)):
- Atomic number \(Z\) (number of protons) = 1
- Mass number \(A\) (total number of protons and neutrons) = 3
- Therefore, the number of neutrons \(N\) = \(A - Z = 3 - 1 = 2\)
So tritium (\( \, ^3_1 H \)) has:
- Protons = 1
- Neutrons = 2
2. Determine the Composition of Helium-3 (\( \, ^3_2 He \)):
- Atomic number \(Z\) = 2
- Mass number \(A\) = 3
- Therefore, the number of neutrons \(N\) = \(A - Z = 3 - 2 = 1\)
So helium-3 (\( \, ^3_2 He \)) has:
- Protons = 2
- Neutrons = 1
3. Balancing Protons and Neutrons in the Decay Equation:
In tritium (\( \, ^3_1 H \)):
- Protons = 1
- Neutrons = 2
In helium-3 (\( \, ^3_2 He \)):
- Protons = 2
- Neutrons = 1
To balance the numbers of protons and neutrons post-decay, we need to add a particle that accounts for the difference. Comparing the initial and final states:
- Protons: Initial = 1 (tritium), Final = 2 (helium-3)
- Neutrons: Initial = 2 (tritium), Final = 1 (helium-3)
We see that:
- Protons have increased by 1 (from 1 to 2)
- Neutrons have decreased by 1 (from 2 to 1)
4. Determine the Particle:
The particle that can be added to balance the equations must therefore:
- Decrease one proton (Initial is 1, final should be 1 meaning one proton participates in increasing another count by 1)
- Increase one neutron (iness doubled the protons and neutrons count resulted in decreased by one )
The particle needs to have:
- Atomic number (protons) = -1
- Mass number (neutrons) = 0
This is characteristic of a beta-minus particle (electron). Therefore the correct particle is:
- Symbol: \( { }_{-1}^0 e \)
The correct answer is:
A. [tex]\( { }_{-1}^0 e \)[/tex]
1. Determine the Composition of Tritium (\( \, ^3_1 H \)):
- Atomic number \(Z\) (number of protons) = 1
- Mass number \(A\) (total number of protons and neutrons) = 3
- Therefore, the number of neutrons \(N\) = \(A - Z = 3 - 1 = 2\)
So tritium (\( \, ^3_1 H \)) has:
- Protons = 1
- Neutrons = 2
2. Determine the Composition of Helium-3 (\( \, ^3_2 He \)):
- Atomic number \(Z\) = 2
- Mass number \(A\) = 3
- Therefore, the number of neutrons \(N\) = \(A - Z = 3 - 2 = 1\)
So helium-3 (\( \, ^3_2 He \)) has:
- Protons = 2
- Neutrons = 1
3. Balancing Protons and Neutrons in the Decay Equation:
In tritium (\( \, ^3_1 H \)):
- Protons = 1
- Neutrons = 2
In helium-3 (\( \, ^3_2 He \)):
- Protons = 2
- Neutrons = 1
To balance the numbers of protons and neutrons post-decay, we need to add a particle that accounts for the difference. Comparing the initial and final states:
- Protons: Initial = 1 (tritium), Final = 2 (helium-3)
- Neutrons: Initial = 2 (tritium), Final = 1 (helium-3)
We see that:
- Protons have increased by 1 (from 1 to 2)
- Neutrons have decreased by 1 (from 2 to 1)
4. Determine the Particle:
The particle that can be added to balance the equations must therefore:
- Decrease one proton (Initial is 1, final should be 1 meaning one proton participates in increasing another count by 1)
- Increase one neutron (iness doubled the protons and neutrons count resulted in decreased by one )
The particle needs to have:
- Atomic number (protons) = -1
- Mass number (neutrons) = 0
This is characteristic of a beta-minus particle (electron). Therefore the correct particle is:
- Symbol: \( { }_{-1}^0 e \)
The correct answer is:
A. [tex]\( { }_{-1}^0 e \)[/tex]
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