Expand your knowledge base with the help of IDNLearn.com's extensive answer archive. Ask anything and receive prompt, well-informed answers from our community of knowledgeable experts.
Sagot :
To determine the particle that completes the given nuclear equation:
[tex]\[ {}_{94}^{239} Pu + {}_0^1 n \rightarrow {}_{40}^{100} Zr + ? + 2 \cdot {}_0^1 n \][/tex]
we need to follow these steps:
1. Calculate the total mass number and atomic number before and after the reaction:
- Before the reaction:
- Plutonium ([tex]\( {}_{94}^{239} Pu \)[/tex]): Mass number is 239, atomic number is 94.
- Neutron ([tex]\( {}_0^1 n \)[/tex]): Mass number is 1, atomic number is 0.
- Total mass number before reaction = 239 (Pu) + 1 (neutron) = 240.
- Total atomic number before reaction = 94 (Pu) + 0 (neutron) = 94.
- After the reaction:
- Zirconium ([tex]\( {}_{40}^{100} Zr \)[/tex]): Mass number is 100, atomic number is 40.
- Two neutrons ([tex]\( 2 \cdot {}_0^1 n \)[/tex]): Mass number is [tex]\( 2 \cdot 1 = 2 \)[/tex], atomic number is [tex]\( 2 \cdot 0 = 0 \)[/tex].
- Let's denote the unknown particle with its mass number [tex]\( A \)[/tex] and atomic number [tex]\( Z \)[/tex].
- Total mass number after the reaction = 100 (Zr) + 2 (from 2 neutrons) + [tex]\( A \)[/tex].
- Total atomic number after the reaction = 40 (Zr) + 0 (from 2 neutrons) + [tex]\( Z \)[/tex].
2. Establish equation for conservation of mass and charge:
- Using conservation of mass number:
[tex]\[ 240 = 100 + 2 + A \implies A = 240 - 102 \implies A = 138 \][/tex]
- Using conservation of atomic number:
[tex]\[ 94 = 40 + 0 + Z \implies Z = 94 - 40 \implies Z = 54 \][/tex]
So, we deduced that the mass number [tex]\( A \)[/tex] of the unknown particle is 138 and the atomic number [tex]\( Z \)[/tex] is 54.
3. Determine the identity of the particle:
- A particle with atomic number [tex]\( Z = 54 \)[/tex] and mass number [tex]\( A = 138 \)[/tex] is Xenon (Xe).
4. Compare with the given options to find the correct one:
- [tex]\({ }_{54}^{138} Xe\)[/tex]
- [tex]\({ }_{54}^{139} Xe\)[/tex]
- [tex]\({ }_{54}^{119} Xe\)[/tex]
- [tex]\({ }_4^{138} Te\)[/tex]
Therefore, the correct option is [tex]\({ }_{54}^{138} Xe\)[/tex].
[tex]\[ {}_{94}^{239} Pu + {}_0^1 n \rightarrow {}_{40}^{100} Zr + ? + 2 \cdot {}_0^1 n \][/tex]
we need to follow these steps:
1. Calculate the total mass number and atomic number before and after the reaction:
- Before the reaction:
- Plutonium ([tex]\( {}_{94}^{239} Pu \)[/tex]): Mass number is 239, atomic number is 94.
- Neutron ([tex]\( {}_0^1 n \)[/tex]): Mass number is 1, atomic number is 0.
- Total mass number before reaction = 239 (Pu) + 1 (neutron) = 240.
- Total atomic number before reaction = 94 (Pu) + 0 (neutron) = 94.
- After the reaction:
- Zirconium ([tex]\( {}_{40}^{100} Zr \)[/tex]): Mass number is 100, atomic number is 40.
- Two neutrons ([tex]\( 2 \cdot {}_0^1 n \)[/tex]): Mass number is [tex]\( 2 \cdot 1 = 2 \)[/tex], atomic number is [tex]\( 2 \cdot 0 = 0 \)[/tex].
- Let's denote the unknown particle with its mass number [tex]\( A \)[/tex] and atomic number [tex]\( Z \)[/tex].
- Total mass number after the reaction = 100 (Zr) + 2 (from 2 neutrons) + [tex]\( A \)[/tex].
- Total atomic number after the reaction = 40 (Zr) + 0 (from 2 neutrons) + [tex]\( Z \)[/tex].
2. Establish equation for conservation of mass and charge:
- Using conservation of mass number:
[tex]\[ 240 = 100 + 2 + A \implies A = 240 - 102 \implies A = 138 \][/tex]
- Using conservation of atomic number:
[tex]\[ 94 = 40 + 0 + Z \implies Z = 94 - 40 \implies Z = 54 \][/tex]
So, we deduced that the mass number [tex]\( A \)[/tex] of the unknown particle is 138 and the atomic number [tex]\( Z \)[/tex] is 54.
3. Determine the identity of the particle:
- A particle with atomic number [tex]\( Z = 54 \)[/tex] and mass number [tex]\( A = 138 \)[/tex] is Xenon (Xe).
4. Compare with the given options to find the correct one:
- [tex]\({ }_{54}^{138} Xe\)[/tex]
- [tex]\({ }_{54}^{139} Xe\)[/tex]
- [tex]\({ }_{54}^{119} Xe\)[/tex]
- [tex]\({ }_4^{138} Te\)[/tex]
Therefore, the correct option is [tex]\({ }_{54}^{138} Xe\)[/tex].
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.
