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Chemistry Question on Bohr’s Model for Hydrogen Atom

The ionization enthalpy of He+ionis19.60×10−18Jatom−1.He^{+} ion is 19.60 \times10^{-18} J atom^{-1}. The ionization enthalpy of Li2+Li^{2+} ion will be

A

84.2×10−18Jatom−184.2\times10^{-18} J atom^{-1}

B

44.10×10−18Jatom−144.10\times10^{-18} J atom^{-1}

C

63.20×10−18Jatom−163.20\times10^{-18} J atom^{-1}

D

21.20×10−18Jatom−121.20\times10^{-18} J atom^{-1}

Answer

44.10×10−18Jatom−144.10\times10^{-18} J atom^{-1}

Explanation

Solution

Ionisation enthalpy ∝Z2(Z=Atomic numder)\propto Z^{2} \left(Z=Atomic\, numder\right) Ionisation enthalpy of Li2+=I.E.ofHe+×(3222)Li^{2+}=I.E. of He^{+}\times\left(\frac{3^{2}}{2^{2}}\right) I.E.of Li2+=19.6×10−18×94=44.1×10−18J atom−1I.E. of \, Li^{2+}=19.6\times10^{-18}\times\frac{9}{4}=44.1\times10^{-18} J \, atom^{-1}