Practice question
Question
The ionization energy of Li²⁺ is 122.4 eV. What is the wavelength of light required to excite an electron from n = 1 to n = 3 in Li²⁺? (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J)
Explanation
ΔE = 108.8 eV. λ ≈ 11.42 nm.
Discussion
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