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For each radionuclide, we will calculate the neutron to proton ratio. We can do that by subtracting the atomic number (number of protons) from the mass number (number of protons + neutrons). (a) \({ }_{5}^{8} \mathrm{~B}\): Neutrons = 8 - 5 = 3; n/p ratio = 3/5 (b) \({ }_{29}^{68} \mathrm{Cu}\): Neutrons = 68 - 29 = 39; n/p ratio = 39/29 (c) Phosphorus-32 (\({ }_{15}^{32} \mathrm{~P}\)): Neutrons = 32 - 15 = 17; n/p ratio = 17/15 (d) Chlorine-39 (\({ }_{17}^{39} \mathrm{~Cl}\)): Neutrons = 39 - 17 = 22; n/p ratio = 22/17

Now, we use the neutron-to-proton ratio to determine the type of radioactive decay. (a) \({ }_{5}^{8} \mathrm{~B}\): n/p ratio is 3/5, which is less than 1, meaning that there are more protons than neutrons. Since the element is light and the atomic number is less than 83, we can expect beta-plus decay (emission of a positron) or electron capture. (b) \({ }_{29}^{68} \mathrm{Cu}\): n/p ratio is 39/29, which is greater than 1, meaning that there are more neutrons than protons. Since the atomic number is less than 83, we can expect beta-minus decay (emission of an electron). (c) Phosphorus 32 (\({ }_{15}^{32} \mathrm{~P}\)): n/p ratio is 17/15, which is slightly greater than 1, meaning that there are a few more neutrons than protons. Since the atomic number is less than 83, we can expect beta-minus decay (emission of an electron). (d) Chlorine-39 (\({ }_{17}^{39} \mathrm{~Cl}\)): n/p ratio is 22/17, which is slightly greater than 1, meaning that there are a few more neutrons than protons. Since the atomic number is less than 83, we can expect beta-minus decay (emission of an electron). In conclusion, the types of radioactive decay for the given radionuclides are: (a) Beta-plus decay or electron capture, (b) Beta-minus decay, (c) Beta-minus decay, (d) Beta-minus decay.