Final answer:
The isotope with 47 protons, 47 electrons, and 60 neutrons is silver-107, symbolized as 107 Ag, making answer option a correct.
Step-by-step explanation:
To identify the isotope with 47 protons, 47 electrons, and 60 neutrons, we look at the atomic number which is equal to the number of protons. The atomic number is the identity of an element and for 47 protons, the element is silver (Ag). The mass number of an isotope is the sum of its protons and neutrons, so for this isotope it is 47 (protons) + 60 (neutrons) = 107. The isotope in question is therefore silver-107, and its symbol is 107 Ag. Since it has an equal number of protons and electrons, it is neutral and not an ion, eliminating option d. 107 Ag+ would indicate a positively charged silver ion with fewer electrons than protons.