Answer
The equation that shows the process for the first ionization energy of lead is $$Pb\to Pb^++e^-$$
The equation that shows the process for the second ionization energy of lead is $$Pb^+\to Pb^{2+}+e^-$$
Work Step by Step
The first ionization is the process to remove the first electron from a neutral lead $(Pb)$ atom. Therefore, the equation that shows the process for the first ionization energy of lead is $$Pb\to Pb^++e^-$$
Similarly, the second ionization is the process to remove another electron from the lead ion, which has already lost an electron before. Therefore, the equation that shows the process for the second ionization energy of lead is $$Pb^+\to Pb^{2+}+e^-$$
