Question

The energy of electron in 1st orbit of Hydrogen atom is –2.18 × 10^-18 Joule. The third ionisation energy of Li⁺² ion is

• (A) 8.72 × 10^-18 J
• (B) 4.36 × 10^-18 J
• (C) 19.62 × 10^-18 J
• (D) 6.54 × 10^-18 J

Answer 1

The Correct Option is C

Explanation:
III lonisation energy of ${\mathrm{Li}}^{+2}$ $=\frac{Z^2}{n^2}{E}_1{H}^{\prime }$ where $(Z=3,n=1)$$=-\frac{Z^2\times 2.18\times {10}^{-18}}{n^2}$ Jule$n=1$ for third ionisation.$=-(-9\times 2.18\times {10}^{-18})$$=19.62\times {10}^{-18}\mathrm{J}$