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 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)1.9 eV,660 nm B)1.9 eV,1050 nm C)2.3 eV,550 nm D)2.3 eV,660 nm E)2.3 eV,1050 nm

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Quizplus · Accepted Answer

The energy difference between n=3 and n=2 levels in hydrogen is 1.9 eV. Using E = hc/λ, the wavelength is calculated to be approximately 660 nm.

A Photon Is Emitted from a Hydrogen Atom That Undergoes