Practice question
Question
The ionization energy of a hydrogen atom is 13.6 eV. What is the wavelength of the photon emitted when an electron in B⁴⁺ falls from n = 5 to n = 2? (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J)
Explanation
For B⁴⁺ (Z=5), ΔE = 71.4 eV. λ ≈ 17.41 nm.
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