To identify which of the given reactions represents a gamma emission, we need to understand the different types of emissions:
1. Alpha Emission (α): Involves the release of an alpha particle ([tex]\(_{2}^{4}He\)[/tex]), leading to a decrease in the atomic number by 2 and the mass number by 4.
2. Beta Emission (β): Involves the transformation of a neutron into a proton with the emission of an electron ([tex]\(\beta\)[/tex]). This increases the atomic number by 1 while the mass number remains unchanged.
3. Gamma Emission (γ): Involves the emission of a gamma photon. This does not alter the atomic number or the mass number of the nucleus.
Now, let's analyze each choice:
A. [tex]\(\mathbf{_{28}^{60} Ni \rightarrow{ }_{28}^{60} Ni + \gamma}\)[/tex]
- Here, the atomic number (28) and mass number (60) of [tex]\(_{28}^{60} Ni\)[/tex] remain unchanged.
- This represents a gamma emission ([tex]\(\gamma\)[/tex]) since it only emits a gamma photon.
B. [tex]\(\mathbf{_{86}^{220} Rn \rightarrow{ }_{84}^{216} Po + { }_{2}^{4} He}\)[/tex]
- Here, [tex]\(_{86}^{220} Rn\)[/tex] transforms into [tex]\(_{84}^{216} Po\)[/tex] and an alpha particle ([tex]\(_{2}^{4} He\)[/tex]) is emitted.
- This represents an alpha emission.
C. [tex]\(\mathbf{_{89}^{228} Ac \rightarrow{ }_{90}^{228} Th + \beta}\)[/tex]
- Here, [tex]\(_{89}^{228} Ac\)[/tex] transforms into [tex]\(_{90}^{228} Th\)[/tex] with the emission of a beta particle ([tex]\(\beta\)[/tex]).
- This represents a beta emission.
D. [tex]\(\mathbf{_{83}^{212} Bi \rightarrow{ }_{84}^{212} Po + { }_{-1}^{0} e}\)[/tex]
- Here, [tex]\(_{83}^{212} Bi\)[/tex] transforms into [tex]\(_{84}^{212} Po\)[/tex] with the emission of a positron ([tex]\(_{-1}^{0} e\)[/tex]).
- This represents another type of beta emission (beta-plus emission or positron emission).
After analyzing each option, we can conclude that:
The reaction that represents a gamma emission is:
A. [tex]\(\mathbf{_{28}^{60} Ni \rightarrow{ }_{28}^{60} Ni + \gamma}\)[/tex]
