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JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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Application +4 | -1
The energies of the electron in a hydrogen like ion are given by :
En=−(2.18×10−18)Z2(1n2)E_n=-\left(2.18 \times 10^{-18}\right) Z^2\left(\frac{1}{n^2}\right), where nn is principal quantum number and ZZ is the atomic number of the element. The ionisation energy (in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} ) of the Li2+\mathrm{Li}^{2+} ion is:
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Correct option: D
11.8×10311.8 \times 10^3
The correct answer is 11.8×10311.8 \times 10^3 because it is calculated using the given formula for ionization energy with Z=3 and n=1, then converting to kJ/mol.

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