Consider the Li^2+ ion (Z=3) in the Bohr model, with its single electron in the stated state. Here a0 is the Bohr radius and the ground-state energy…

In the Bohr model $r_n=\dfrac{n^2a_0}{Z}$ and $E_n=-13.6\,\dfrac{Z^2}{n^2}\ \mathrm{eV}$. (P) $r_2=\dfrac{4a_0}{3}$, which is (4). (Q) $E_3=-13.6\times\dfrac{9}{9}=-13.6\ \mathrm{eV}$, which is (3). (R) The orbit circumference holds $n$ de Broglie wavelengths, $2\pi r_n=n\lambda$, so $\lambda=\dfrac{2\pi r_3}{3}=\dfrac{2\pi}{3}\cdot\dfrac{9a_0}{3}=2\pi a_0$, which is (1). (S) Removing the electron from $n=2$ needs $|E_2|=13.6\times\dfrac94=30.6\ \mathrm{eV}$, which is (5). The $122.4\ \mathrm{eV}$ in (2) is the ground-state ionisation energy and is a distractor. All four entries follow from $r_n=n^2a_0/Z$, $E_n=-13.6Z^2/n^2$ and the quantisation of the orbit circumference in de Broglie wavelengths. Hence P→4, Q→3, R→1, S→5.