Answer :
Let's solve the given alpha decay reaction step-by-step.
The reaction is:
[tex]\[ {}_{98}^{251} Cf \rightarrow {}_{96}^{247} Cm + \text{?} \][/tex]
1. Understanding Alpha Decay:
- Alpha decay is a type of radioactive decay where an alpha particle is emitted from a nucleus.
- An alpha particle is a helium-4 nucleus, denoted as \({}_2^4 He\).
- When an alpha particle is emitted, the parent nucleus loses 2 protons and 2 neutrons. This means:
- The atomic number decreases by 2.
- The mass number decreases by 4.
2. Determine the Change in Atomic Number and Mass Number:
- Initial atomic number of Californium (\(\text{Cf}\)): 98
- Final atomic number of Curium (\(\text{Cm}\)): 96
- Change in atomic number: \(98 - 96 = 2\)
- Initial mass number of Californium (\(\text{Cf}\)): 251
- Final mass number of Curium (\(\text{Cm}\)): 247
- Change in mass number: \(251 - 247 = 4\)
3. Comparison with Alpha Particle:
- An alpha particle has an atomic number of 2 and a mass number of 4.
- The changes we calculated (2 in atomic number and 4 in mass number) match those of an alpha particle.
4. Conclusion:
- The missing particle in the reaction must be an alpha particle \({}_2^4 He\).
Therefore, the correct answer is:
[tex]\[ \boxed{{}_2^4 He} \][/tex]
This corresponds to option C in the given choices.
Thus, the correct answer is:
C. [tex]\({}_2^4 He\)[/tex]
The reaction is:
[tex]\[ {}_{98}^{251} Cf \rightarrow {}_{96}^{247} Cm + \text{?} \][/tex]
1. Understanding Alpha Decay:
- Alpha decay is a type of radioactive decay where an alpha particle is emitted from a nucleus.
- An alpha particle is a helium-4 nucleus, denoted as \({}_2^4 He\).
- When an alpha particle is emitted, the parent nucleus loses 2 protons and 2 neutrons. This means:
- The atomic number decreases by 2.
- The mass number decreases by 4.
2. Determine the Change in Atomic Number and Mass Number:
- Initial atomic number of Californium (\(\text{Cf}\)): 98
- Final atomic number of Curium (\(\text{Cm}\)): 96
- Change in atomic number: \(98 - 96 = 2\)
- Initial mass number of Californium (\(\text{Cf}\)): 251
- Final mass number of Curium (\(\text{Cm}\)): 247
- Change in mass number: \(251 - 247 = 4\)
3. Comparison with Alpha Particle:
- An alpha particle has an atomic number of 2 and a mass number of 4.
- The changes we calculated (2 in atomic number and 4 in mass number) match those of an alpha particle.
4. Conclusion:
- The missing particle in the reaction must be an alpha particle \({}_2^4 He\).
Therefore, the correct answer is:
[tex]\[ \boxed{{}_2^4 He} \][/tex]
This corresponds to option C in the given choices.
Thus, the correct answer is:
C. [tex]\({}_2^4 He\)[/tex]
