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Question

Ionization energy of $${He}^{+}$$ is $$19.6 \times {10}^{-18} J {atom}^{-1}$$. The energy of the first stationary state $$\left(n=1\right)$$ of $${Li}^{2+}$$ is:


A
2.2×1015Jatom1
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B
8.82×1017Jatom1
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C
4.41×1016Jatom1
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D
4.41×1017Jatom1
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Solution

The correct option is D $$-4.41 \times {10}^{-17} J {atom}^{-1}$$
$$\dfrac{{I}_{{He}^{+}}}{{I}_{{Li}^{2+}}}=\dfrac{{Z}_{1}^{2}}{{Z}_{2}^{2}}$$

$$\dfrac{19.6 \times {10}^{-18}}{{I}_{{Li}^{2+}}}=\dfrac{4}{9}$$

$${I}_{{Li}^{2+}}=\dfrac{9}{4} \times 19.6 \times {10}^{-18} = 44.1 \times {10}^{-18}$$
         $$=4.41 \times {10}^{-17} J {atom}^{-1}$$

$${E}_{{Li}^{2+}}=-4.41 \times {10}^{-17} J {atom}^{-1}$$

Chemistry

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