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Ionisation energy of $\mathrm{He}^{+}$is $19.6 \times 10^{-18} \mathrm{~J}$ atom $^{-1}$. The energy of the first stationary state $(n=1)$ of $\mathrm{Li}^{2+}$ is
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The correct answer is:
$-4.41 \times 10^{-17} \mathrm{~J}$ atom $^{-1}$
I. $E=\frac{Z^{2}}{n^{2}} \times 13.6 \mathrm{eV}....(i)$
or $\frac{I_{1}}{I_{2}}=\frac{Z_{1}^{2}}{n_{1}^{2}} \times \frac{n_{2}^{2}}{Z_{2}^{2}}....(ii)$
Given $I_{1}=-19.6 \times 10^{-18}, Z_{1}=2$, $n_{1}=1, Z_{2}=3$ and $n_{2}=1$
Substituting these values in equation (ii).
$$
-\frac{19.6 \times 10^{-18}}{\mathrm{I}_{2}}=\frac{4}{1} \times \frac{1}{9}
$$
or $I_{2}=-19.6 \times 10^{-18} \times \frac{9}{4}$ $=-4.41 \times 10^{-17} \mathrm{~J} /$ atom
or $\frac{I_{1}}{I_{2}}=\frac{Z_{1}^{2}}{n_{1}^{2}} \times \frac{n_{2}^{2}}{Z_{2}^{2}}....(ii)$
Given $I_{1}=-19.6 \times 10^{-18}, Z_{1}=2$, $n_{1}=1, Z_{2}=3$ and $n_{2}=1$
Substituting these values in equation (ii).
$$
-\frac{19.6 \times 10^{-18}}{\mathrm{I}_{2}}=\frac{4}{1} \times \frac{1}{9}
$$
or $I_{2}=-19.6 \times 10^{-18} \times \frac{9}{4}$ $=-4.41 \times 10^{-17} \mathrm{~J} /$ atom
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