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The ionisation energy of He⁺ is 19.6 × 10⁻¹⁸ J atom⁻¹. What is the energy of the first stationary state (n = 1) of Li²⁺?
- 4.41 × 10⁻¹⁶ J atom⁻¹
- 4.41 × 10⁻¹⁷ J atom⁻¹
- −2.2 × 10⁻¹⁵ J atom⁻¹
- 8.82 × 10⁻¹⁷ J atom⁻¹
Correct answer: 4.41 × 10⁻¹⁷ J atom⁻¹
Solution
E(Li2+, n=1) = E(He+, n=1) x (3/2)^2 = 19.6e-18 x 2.25 = 4.41e-17 J/atom. The stored value 4.41e-16 is ten times too large.
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