The ratio of ground state energy of $\mathrm{Li}^{2+}, \mathrm{He}^{+}, \mathrm{H}$ is
The ratio of ground state energy of $\mathrm{Li}^{2+}, \mathrm{He}^{+}, \mathrm{H}$ is
$3: 2: 1$
$1: 2: 3$
$9: 4: 1$
$1: 4: 9$
Solution
Ground state energy of hydrogen like atom is
$E=-13.6 \times \frac{Z^2}{n^2}$
All are in ground state i.e $\mathrm{n}=1$
$\therefore E \propto Z^2$
$\therefore \quad$ The ratio of ground state energy of
$\begin{aligned}
& \mathrm{E}_{\mathrm{Li}^{2+}}: \mathrm{E}_{\mathrm{He}^{+}}: \mathrm{E}_{\mathrm{H}}(3)^2:(2)^2:(1)^2 \\
&=9: 4: 1
\end{aligned}$