Sure, let's write the balanced nuclear equations for each of the specified processes.
### (a) Alpha emission from curium-242:
In alpha decay, the original nucleus loses 2 protons and 2 neutrons (an alpha particle, [tex]\( \alpha \)[/tex] or [tex]\( \text{ }_2^4 \text{He} \)[/tex]).
The original element is curium-242:
[tex]\[ \mathrm{^{242}_{96}Cm} \][/tex]
After emitting an alpha particle ([tex]\( \mathrm{^4_2He} \)[/tex]):
[tex]\[ \mathrm{^{242}_{96}Cm \rightarrow \mathrm{^4_2He} + \,^{238}_{94}X} \][/tex]
So, the resulting nucleus will have:
- Atomic number: [tex]\( 96 - 2 = 94 \)[/tex]
- Mass number: [tex]\( 242 - 4 = 238 \)[/tex]
The element with atomic number 94 is plutonium ([tex]\( \mathrm{Pu} \)[/tex]).
Therefore, the balanced equation is:
[tex]\[ \mathrm{^{242}_{96}Cm \rightarrow \mathrm{^4_2He} + \,^{238}_{94}Pu} \][/tex]
### (b) Beta emission from magnesium-28:
In beta decay (beta-minus decay), a neutron is transformed into a proton, and an electron (beta particle, [tex]\( \beta^- \)[/tex] or [tex]\( \mathrm{_{-1}e} \)[/tex]) is emitted.
The original element is magnesium-28:
[tex]\[ \mathrm{^{28}_{12}Mg} \][/tex]
After emitting a beta particle ([tex]\( \mathrm{_{-1}e} \)[/tex]):
[tex]\[ \mathrm{^{28}_{12}Mg \rightarrow \,^{28}_{13}X + \mathrm{_{-1}e}} \][/tex]
So, the resulting nucleus will have:
- Atomic number: [tex]\( 12 + 1 = 13 \)[/tex]
- Mass number: [tex]\( 28 \)[/tex] (remains the same)
The element with atomic number 13 is aluminum ([tex]\( \mathrm{Al} \)[/tex]).
Therefore, the balanced equation is:
[tex]\[ \mathrm{^{28}_{12}Mg \rightarrow \,^{28}_{13}Al + \mathrm{_{-1}e}} \][/tex]
### (c) Positron emission from xenon-118:
In positron emission (beta-plus decay), a proton is transformed into a neutron, and a positron (beta-plus particle, [tex]\( \beta^+ \)[/tex] or [tex]\( \mathrm{^0_{+1}e} \)[/tex]) is emitted.
The original element is xenon-118:
[tex]\[ \mathrm{^{118}_{54}Xe} \][/tex]
After emitting a positron ([tex]\( \mathrm{^0_{+1}e} \)[/tex]):
[tex]\[ \mathrm{^{118}_{54}Xe \rightarrow \,^{118}_{53}X + \mathrm{^0_{+1}e}} \][/tex]
So, the resulting nucleus will have:
- Atomic number: [tex]\( 54 - 1 = 53 \)[/tex]
- Mass number: [tex]\( 118 \)[/tex] (remains the same)
The element with atomic number 53 is iodine ([tex]\( \mathrm{I} \)[/tex]).
Therefore, the balanced equation is:
[tex]\[ \mathrm{^{118}_{54}Xe \rightarrow \,^{118}_{53}I + \mathrm{^0_{+1}e}} \][/tex]
