Question

The ratio of ground state energy of $L{i}^{2+},H{e}^{+}$ and H is :

• A. 1 : 2 : 3
• B. 3 : 2 : 1
• C. 1 : 4 : 9
• D. 9 : 4 : 1

Answer 1

The correct option is D 9 : 4 : 1

$E_n$ = - $R_H$ ( $Z^2/n^2$)

$E_{L{i}^{2+}}$ = - $R_H$ ( $3^2/1^2$), since, $Z=3,n=1$.
$E_{L{i}^{2+}}$ = - $R_H$ ( $9$ )

$E_{H{e}^{+}}$ = - $R_H$ ( $2^2/1^2$)
$E_{H{e}^{+}}$ = - $R_H$ ( $4$ )

$E_H$ = - $R_H$ ( $1^2/1^2$)
$E_H$ = - $R_H$ ( $1$ )

The ratio of ground state energy of $L{i}^{2+},H{e}^{+}$ and H is $9:4:1$.