Question

The ionization potential of Li⁺⁺ ion is 122.4 V. Then, its first excitation potential is –

• A. 122.4 V
• B. 91.8 V
• C. 61.2 V
• D. 30.6 V

Answer 1

Answer: (B)

91.8 V

Answer 2

∆V = V₂ – V₁ = 13.6 × 3²$\left( \frac { 1 } { 1 ^ { 2 } } - \frac { 1 } { 2 ^ { 2 } } \right)$

= 13.6 × 9 × $\frac { 3 } { 4 }$