# Chapter 5 - Nuclear Chemistry - 5.2 Nuclear Reactions - Questions and Problems - Page 162: 5.20

a) $^{11}_6$C --> $^{11}_5$B + $^0_{+1}$e ; Positron Emission b) $^{35}_{16}$S -->$^{35}_{17}$Cl + $^0_{-1}$e ; Beta Decay c) $^{90}_{38}$Sr --> $^{90}_{39}$Y + $^0_{-1}e$ ; Beta Decay d) $^{210}_{83}$Bi --> $^{206}_{81}$Tl + $^4_2$He ; Alpha Decay e) $^{89}_{40}$Zr --> $^{89}_{39}$Y + $^0_{+1}$e ; Positron Emission

#### Work Step by Step

a) The atomic number went down by one, but the mass number remained the same, so the missing particle is a Positron ($^0_{+1}$e), thus making it a Positron Emission reaction. b) There was a Beta Particle (negative charge) emitted on the other side of the equation, so the resulting isotope would have an atomic number of +1 to the original, which is indicative of Beta Decay. c) The product side of the equation shows $^{90}_{39}$Y + $^0_{-1}e$ which means the original isotope had an atomic number that was one less than the product. The presence of a Beta Particle is indicative of a Beta Decay reaction. d) An Alpha Particle ($^4_2$He) is part of the product, which is indicative of an Alpha Decay reaction. To balance the equation and find the missing isotope, we must subtract 4 from the mass number and 2 from the atomic number of the original isotope. e) The product in this reaction is a Positron and an isotope, which means this was a Positron Emission reaction. To balance to equation the original isotope will have the same mass number but one must be added to the atomic number of the product isotope.

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