To determine the type of nuclear reaction in the equation [tex]\(_{62}^{146} Sm \rightarrow _{60}^{142} Nd + _2^4 He\)[/tex], let's break it down step by step:
1. Identify the elements involved:
- The initial element is Samarium ([tex]\( Sm \)[/tex]) with atomic number 62 and mass number 146.
- The resulting elements are Neodymium ([tex]\( Nd \)[/tex]) with atomic number 60 and mass number 142, and Helium ([tex]\( He \)[/tex]) with atomic number 2 and mass number 4.
2. Examine the changes in atomic and mass numbers:
- Atomic number (number of protons):
- Initially, the atomic number is 62.
- After the reaction, we have Neodymium with an atomic number of 60 and Helium with an atomic number of 2.
- Adding these, we get [tex]\( 60 + 2 = 62 \)[/tex], which is equal to the original atomic number.
- Mass number (total number of protons and neutrons):
- Initially, the mass number is 146.
- After the reaction, the mass numbers are Neodymium (142) and Helium (4).
- Adding these, we get [tex]\( 142 + 4 = 146 \)[/tex], which matches the original mass number.
3. Determine the type of emitted particle:
- The reaction results in the emission of a Helium nucleus, denoted by [tex]\( _2^4 He \)[/tex].
- This particle consists of 2 protons and 2 neutrons, which is characteristic of an alpha particle.
Given this information, the reaction involves the emission of an alpha particle.
Therefore, the nuclear reaction in the equation [tex]\(_{62}^{146} Sm \rightarrow _{60}^{142} Nd + _2^4 He\)[/tex] is:
C. Alpha decay
