Answer: 4. [tex]^{0}_{-1}\textrm{e}[/tex] and [tex]^{0}_{1}\textrm{e}[/tex]
Explanation:
a) The given reaction is [tex]^{41}_{20}\textrm{Ca}+^{x}_{y}\textrm{X}\rightarrow ^{41}_{19}\textrm {K}[/tex]
As the mass on both reactant and product side must be equal:
[tex]41+x=41[/tex]
[tex]x=0[/tex]
As the atomic number on both reactant and product side must be equal:
[tex]20+y=19[/tex]
[tex]y=-1[/tex]
[tex]^{41}_{20}\textrm{Ca}+^{0}_{-1}\textrm{e}\rightarrow ^{41}_{19}\textrm {K}[/tex]
b) [tex]^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{x}_{y}\textrm{X}[/tex]
Total mass on reactant side = total mass on product side
15 =15 + x
x = 0
Total atomic number on reactant side = total atomic number on product side
8 = 7 + y
y = 1
[tex]^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{0}_{1}\textrm{e}[/tex]
