Question

The hydrogen-like species Li${}^{2+}$ is in a spherically symmetric state S${}_1$ with one radial node and its energy is equal to the ground state energy of the hydrogen atom.

The energy of the state S${}_2$ (next state) in units of the hydrogen atom ground state energy is:

• A. $7.5$
• B. $1.50$
• C. $2.25$
• D. $4.50$

Answer 1

The correct option is C $2.25$

S${}_1$ state spherical symmetrical means l = 0,Number of nodal points = 1 (n - l - 1) = 1
(n - 0 - 1) = 1 $\therefore$ n = 2
Hence 2s orbital corresponds to S${}_1$ state
S${}_2$ state has only one radial node and E = ground state of H atom
E${}_{H_2^{\oplus }}$ = $\frac{R}{1^2}$(for ground state)
E in S${}_1$ state Li${}^{2+}$ = -$\frac{R}{1^2}$(but this is equal -$\frac{Z^2}{n^2}$ $\times$ $R$)
Since Z = 3 for Li${}^{2+}$ ion
$\therefore$ n = 3
Since it has only one radial node:
(n - l - 1) = 1, $\Rightarrow$ (3 - l - 1), $\therefore$ $l=1$
Therefore, S${}_2$ state corresponds to $3p(n=3,l=1)$
E of S${}_2$ state = $\frac{R{Z}^2}{n^2}$ = $R$ $\times$ $\frac{3^2}{2^2}$ = $\frac{9}{4}$R = $2.25R$
2.25 $\times$ E${}_{H_2^{\oplus }}$ in ground state