If a singly ionized helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level E f of the atom, and the n value, n f

Question

If a singly ionized helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level E f ​ of the atom, and the n value, n f ​ , of this final state?

TestHub · Accepted Answer

He + at n = 4 : E 4 ​ = − 54.4/16 = − 3.4 eV . Photon: 1240/470 = 2.64 eV . E f ​ = − 3.4 − 2.64 = − 6.04 ≈ − 6.0 eV = − 54.4/9 ⇒ n f ​ = 3.

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