Answer :
To determine which reaction is an example of neutron bombardment, let's analyze each given reaction in detail:
1. [tex]\(\text{${ }_{21}^{46} S c \rightarrow{ }_{21}^{46} S c+$ energy}\)[/tex]:
- This reaction demonstrates an excited atom (Sc-46) returning to a ground state by releasing energy. It does not involve neutron bombardment.
2. [tex]\(\text{${ }_{96}^{242} Cm +{ }_2^4 He \rightarrow{ }_{98}^{245} Cf +\frac{1}{0} n$}\)[/tex]:
- Here, curium-242 ([tex]\( \text{Cm-242} \)[/tex]) reacts with an alpha particle ([tex]\( \text{He-4} \)[/tex]) to form californium-245 ([tex]\( \text{Cf-245} \)[/tex]) and a neutron ([tex]\( \text{n} \)[/tex]). This involves alpha particle bombardment, not neutron bombardment.
3. [tex]\(\text{${ }_{92}^{235} U+{ }_0^1 n \rightarrow{ }_{52}^{138} T e+{ }_{40}^{96} Z r+2{ }_0^1 n$}\)[/tex]:
- In this reaction, uranium-235 ([tex]\( \text{U-235} \)[/tex]) reacts with a neutron ([tex]\( \text{n} \)[/tex]) to produce tellurium-138 ([tex]\( \text{Te-138} \)[/tex]), zirconium-96 ([tex]\( \text{Zr-96} \)[/tex]), and two neutrons ([tex]\( \text{n} \)[/tex]). This is an example of a neutron bombardment reaction.
4. [tex]\(\text{${ }_{96}^{242} Cm \rightarrow{ }_{94}^{238} Pu +{ }_2^4 He$}\)[/tex]:
- This reaction shows curium-242 ([tex]\( \text{Cm-242} \)[/tex]) spontaneously decaying into plutonium-238 ([tex]\( \text{Pu-238} \)[/tex]) and an alpha particle ([tex]\( \text{He-4} \)[/tex]). This is an alpha decay, not neutron bombardment.
5. [tex]\(\text{${ }_{22}^{51} Ti \rightarrow{ }_{23}^{51} V+{ }_{-1}^0 \beta$}\)[/tex]:
- This reaction represents titanium-51 ([tex]\( \text{Ti-51} \)[/tex]) undergoing beta decay, forming vanadium-51 ([tex]\( \text{V-51} \)[/tex]) and a beta particle ([tex]\( \beta \)[/tex]). This does not involve neutron bombardment.
Among the given reactions, the third reaction:
[tex]\[ \text{${ }_{92}^{235} U+{ }_0^1 n \rightarrow{ }_{52}^{138} T e+{ }_{40}^{96} Z r+2{ }_0^1 n$} \][/tex]
is the example of a neutron bombardment reaction.
Thus, the correct answer is the third reaction.
1. [tex]\(\text{${ }_{21}^{46} S c \rightarrow{ }_{21}^{46} S c+$ energy}\)[/tex]:
- This reaction demonstrates an excited atom (Sc-46) returning to a ground state by releasing energy. It does not involve neutron bombardment.
2. [tex]\(\text{${ }_{96}^{242} Cm +{ }_2^4 He \rightarrow{ }_{98}^{245} Cf +\frac{1}{0} n$}\)[/tex]:
- Here, curium-242 ([tex]\( \text{Cm-242} \)[/tex]) reacts with an alpha particle ([tex]\( \text{He-4} \)[/tex]) to form californium-245 ([tex]\( \text{Cf-245} \)[/tex]) and a neutron ([tex]\( \text{n} \)[/tex]). This involves alpha particle bombardment, not neutron bombardment.
3. [tex]\(\text{${ }_{92}^{235} U+{ }_0^1 n \rightarrow{ }_{52}^{138} T e+{ }_{40}^{96} Z r+2{ }_0^1 n$}\)[/tex]:
- In this reaction, uranium-235 ([tex]\( \text{U-235} \)[/tex]) reacts with a neutron ([tex]\( \text{n} \)[/tex]) to produce tellurium-138 ([tex]\( \text{Te-138} \)[/tex]), zirconium-96 ([tex]\( \text{Zr-96} \)[/tex]), and two neutrons ([tex]\( \text{n} \)[/tex]). This is an example of a neutron bombardment reaction.
4. [tex]\(\text{${ }_{96}^{242} Cm \rightarrow{ }_{94}^{238} Pu +{ }_2^4 He$}\)[/tex]:
- This reaction shows curium-242 ([tex]\( \text{Cm-242} \)[/tex]) spontaneously decaying into plutonium-238 ([tex]\( \text{Pu-238} \)[/tex]) and an alpha particle ([tex]\( \text{He-4} \)[/tex]). This is an alpha decay, not neutron bombardment.
5. [tex]\(\text{${ }_{22}^{51} Ti \rightarrow{ }_{23}^{51} V+{ }_{-1}^0 \beta$}\)[/tex]:
- This reaction represents titanium-51 ([tex]\( \text{Ti-51} \)[/tex]) undergoing beta decay, forming vanadium-51 ([tex]\( \text{V-51} \)[/tex]) and a beta particle ([tex]\( \beta \)[/tex]). This does not involve neutron bombardment.
Among the given reactions, the third reaction:
[tex]\[ \text{${ }_{92}^{235} U+{ }_0^1 n \rightarrow{ }_{52}^{138} T e+{ }_{40}^{96} Z r+2{ }_0^1 n$} \][/tex]
is the example of a neutron bombardment reaction.
Thus, the correct answer is the third reaction.