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^-³⁴J.s, c = 3.00 * 10⁸ m/s, 1 eV = 1.60 *10^-¹⁹ J, and 1 nm = 10^-⁹ m) A) 2.21 eV, 563 nm B) 1.89 eV, 460 nm C) 2.58 * 10^-19 J, 658 nm D) 3.02 * 10^-19 J, 658 nm

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Quizplus · Accepted Answer

The energy difference between n=3 and n=2 levels in a hydrogen atom is calculated using the formula E = 13.6 eV * (1/n2^2 - 1/n1^2). This gives approximately 1.89 eV, which converts to about 3.02 x 10^-19 J. Using E = hc/λ, the wavelength λ is calculated to be approximately 658 nm.

A Photon Is Emitted from a Hydrogen Atom That Undergoes