In most multi-electron atoms, the 2s peak appears to the left of (that is, at a higher binding energy than) the 2p peak in a photoelectron spectrum. Which of the following best explains this difference?
a) 2s electrons are more effectively shielded by the inner 1s electrons than 2p electrons, so they experience a smaller effective nuclear charge and therefore require more energy to remove.
b) 2s electrons are located farther from the nucleus than 2p electrons, so a larger amount of energy must be supplied to reach and remove them.
c) 2s electrons penetrate closer to the nucleus and experience less shielding than 2p electrons, so a larger energy is required to remove them.
d) Greater electron–electron repulsion in the 2s subshell raises its binding energy above that of the 2p subshell.