$Li$ has a higher first ionization energy than $Na$ since its outermost electron is more likely to be closer to the nucleus than that of $Na$. Therefore, it undergoes more attraction from the nucleus, making it harder to remove.
Work Step by Step
The first ionization energy is the minimum energy required to remove the first electron from a neutral atom. The amount of energy required to remove an electron depends on whether that electron is strongly attracted to the nucleus or not. In the $Li$ atom, the outermost electron is in the $n=2$ shell (since $Li$ lies in the second row of the periodic table), while in the $Na$ atom, the outermost electron is in the $n=3$ shell (since $Na$ lies in the third row of the periodic table). We know that $n=3$ shell is further from the nucleus than the $n=2$ shell. Therefore, the outermost electron of $Na$ atom has a higher chance of being further from the nucleus than that of $Li$ atom. Being further means there is less attraction from the nucleus. Thus, it is easier to remove the outermost electron of $Na$ atom due to less attraction from the nucleus. The first ionization energy of $Na$, as a result, is lower than that of $Li$.