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
B
8.82×1017Jatom1
C
4.41×1016Jatom1
D
4.41×1017Jatom1

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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