Question

The ionization energy of He${}^{+}$ is 19.6 $\times$ 10${}^{-18}$ J atoms. Calculate the energy of first stationary state of Li${}^{2+}$ :

• A. $12\times {10}^{-18}$J
• B. $4.4\times {10}^{-18}$J
• C. $2.2\times {10}^{-18}$J
• D. None of these

Answer 1

Answer: (B)

$4.4\times {10}^{-18}$J

IE of an ion = IE of H atom $\times$ Z${}^2$

IE of H atom = $\frac{\text{IE of ion}}{Z^2}$ = $\frac{{\text{IE of He}}^{\oplus }}{2^2}$ = $\frac{19.6\times {10}^{-18}}{4}$ J = 4.9 $\times$ 10${}^{-18}$ J

IE of Li${}^{2+}$ = $4.9\times$ 10${}^{-18}$ J $\times$ 3${}^2$ = $4.41\times$ 10${}^{-18}$ J