91Ó°ÊÓ

To determine the number of neutrons, we will use the following formula: neutrons = mass number (A) - atomic number (Z). Using this equation, we can calculate the number of neutrons for each given isotope. (a) For \({ }_{32}^{66} \mathrm{Ge}\): Neutrons = 66 - 32 = 34 (b) For \({ }_{45}^{105} \mathrm{Rh}\): Neutrons = 105 - 45 = 60 (c) For Iodine-137: Neutrons = 137 - 53 (since iodine has an atomic number of 53) = 84 (d) For Cerium-133: Neutrons = 133 - 58 (since cerium has an atomic number of 58) = 75

Now, we will calculate and compare the N/Z ratios of each isotope to determine their stability. (a) For \({ }_{32}^{66} \mathrm{Ge}\): N/Z = 34/32 = 1.0625 (b) For \({ }_{45}^{105} \mathrm{Rh}\): N/Z = 60/45 = 1.3333 (c) For Iodine-137: N/Z = 84/53 = 1.5849 (d) For Cerium-133: N/Z = 75/58 = 1.2931

Based on the N/Z ratios calculated in the previous step, we can predict whether each isotope is likely to undergo beta decay (where a neutron turns into a proton) or positron emission (where a proton turns into a neutron). (a) For \({ }_{32}^{66} \mathrm{Ge}\): The N/Z ratio is slightly above 1, implying a mild neutron excess. This nucleus might prefer to undergo beta decay, as it will increase its proton number by one and decrease its neutron number by one, thus making the ratio closer to 1. (b) For \({ }_{45}^{105} \mathrm{Rh}\): The N/Z ratio is above 1, indicating a neutron excess. This nucleus is likely to undergo beta decay, which will help bring the ratio closer to the ideal 1:1 ratio. (c) For Iodine-137: The N/Z ratio is considerably above 1, suggesting a significant neutron excess. As a result, this nucleus might also prefer to undergo beta decay to increase its stability. (d) For Cerium-133: The N/Z ratio is above 1, also implying neutron excess. Thus, this nucleus might too prefer to undergo beta decay to stabilize its N/Z ratio. Overall, all of the given isotopes are more likely to undergo beta decay, as it will help bring the ratios closer to 1:1, increasing their stability.