Question

If the ionization energy of $H{e}^{+}$ is $19.6\times {10}^{-18}$ J per atom, then the energy of $L{i}^{2+}$ ion in the second stationary state is:

• A. $-4.9\times {10}^{-18}$ J
• B. $-44.1\times {10}^{-18}$ J
• C. $-11.025\times {10}^{-18}$ J
• D. None of these

Answer 1

Answer: (C)

$-11.025\times {10}^{-18}$ J

$E=Z^2\times E_{H^{+}}$

For $H{e}^{+}$ ion, $Z=2$.

Hence, $E_H=E/Z^2=-(19.6\times {10}^{-18}/4)=-4.9\times {10}^{-18}$

For $L{i}^{2+}$ ion, $Z=3$.

Hence, $E=3^2\times E_H=9\times (-4.9\times {10}^{-18})=-44.1\times {10}^{-18}$

Energy of second orbit, $E^{\prime }=E/n^2=-44.1\times {10}^{-18}/2^2=-11.025\times {10}^{-18}J$