JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Atomic Structure

The electron removed when a lithium atom is ionised is its outermost one, which occupies the second shell, so the atom is treated as a one-electron (hydrogen-like) species with nuclear charge $Z=3$ and principal quantum number $n=2$. The binding energy of such an electron scales as $Z^2/n^2$ times the first-shell energy of hydrogen, so its magnitude is $\frac{Z^2}{n^2}\times 2.2\times 10^{-18}=\frac{9}{4}\times 2.2\times 10^{-18}=4.95\times 10^{-18}$ J. The longest wavelength able to ionise the atom corresponds to exactly this minimum energy through $E=hc/\lambda$, giving $\lambda =\frac{hc}{E}=\frac{(6.63\times 10^{-34})(3\times 10^8)}{4.95\times 10^{-18}}\approx 4.0\times 10^{-8}$ m, so $x=4$.