Final answer:
The question asks for the nuclear equations for certain isotopes undergoing alpha decay. Alpha decay results in the emission of an alpha particle (a helium nucleus), reducing the original atom's atomic number by two and mass number by four. The nuclear equations for Radium-226, Polonium-214, Thorium-230, and Polonium-210 have been provided.
Step-by-step explanation:
The question involves writing nuclear equations for the alpha decay of certain radioactive isotopes. Alpha decay is a type of radioactive decay where an unstable nucleus emits an alpha particle, which is essentially a helium nucleus made up of two protons and two neutrons. As a result of this emission, the original atom loses two protons and two neutrons, leading to a decrease in both its atomic number by two and mass number by four. Here are the correct nuclear equations for the alpha decay processes asked in the question:
- Radium-226 (^226Ra₈₈) undergoes alpha decay to form radon-222 (^222Rn₈₆), so the nuclear equation is: ^226Ra₈₈ → ^222Rn₈₆ + ^4He₂
- Polonium-214 (^214Po₈₄) undergoes alpha decay to form lead-210 (^210Pb₈₂), so the nuclear equation is: ^214Po₈₄ → ^210Pb₈₂ + ^4He₂
- Thorium-230 (^230Th₉₀) undergoes alpha decay to form radium-226 (^226Ra₈₈), so the nuclear equation is: ^230Th₉₀ → ^226Ra₈₈ + ^4He₂
- Polonium-210 (^210Po₈₄) undergoes alpha decay to form lead-206 (^206Pb₈₂), so the nuclear equation is: ^210Po₈₄ → ^206Pb₈₂ + ^4He₂