A photon is emitted from a hydrogen atom that undergoes a transition from n = 3 to n = 2.Calculate the energy and wavelength of the photon.(The ionization energy of hydrogen is 13.6 eV,and h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9 m) A)1.89 eV,658 nm B)2.21 eV,563 nm C)1.89 eV,460 nm D)3.19 eV,658 nm

Question

Quizplus · Accepted Answer

The energy difference between n=3 and n=2 levels in a hydrogen atom is 1.89 eV, which corresponds to a wavelength of approximately 658 nm.

A Photon Is Emitted from a Hydrogen Atom That Undergoes