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Ionisation energy of $\mathrm{He}^{+}$is $19.6 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$. The energy of the first stationary state $(\mathrm{n}=1)$ of $\mathrm{Li}^{2+}$ is
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The correct answer is:
$-4.41 \times 10^{-17} \mathrm{~J}_{\text {atom }}{ }^{-1}$
$-4.41 \times 10^{-17} \mathrm{~J}_{\text {atom }}{ }^{-1}$
$$
\mathrm{IE}_{\mathrm{He}^{+}}=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \text { where }\left(\mathrm{Z}_{\mathrm{He}^{+}}=2\right)
$$
Hence $13.6 \times Z_{\mathrm{He}^{+}}^2=19.6 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$.
$$
\left(\mathrm{E}_1\right)_{\mathrm{L}^{+2}}=-13.6 \mathrm{Z}_{\mathrm{Uj}^{+2}}^2 \times \frac{1}{1^2}=-13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \times\left[\frac{\mathrm{Z}_{\mathrm{L}^{+2}}^2}{\mathrm{Z}_{\mathrm{He}^{+}}^2}\right]=-19.6 \times 10^{-18} \times \frac{9}{4}=-4.41 \times 10^{-17} \mathrm{~J} / \text { atom }
$$
\mathrm{IE}_{\mathrm{He}^{+}}=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \text { where }\left(\mathrm{Z}_{\mathrm{He}^{+}}=2\right)
$$
Hence $13.6 \times Z_{\mathrm{He}^{+}}^2=19.6 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$.
$$
\left(\mathrm{E}_1\right)_{\mathrm{L}^{+2}}=-13.6 \mathrm{Z}_{\mathrm{Uj}^{+2}}^2 \times \frac{1}{1^2}=-13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \times\left[\frac{\mathrm{Z}_{\mathrm{L}^{+2}}^2}{\mathrm{Z}_{\mathrm{He}^{+}}^2}\right]=-19.6 \times 10^{-18} \times \frac{9}{4}=-4.41 \times 10^{-17} \mathrm{~J} / \text { atom }
$$
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