For hydrogen,what is the wavelength of the photon emitted when an electron drops from a 3d orbital to a 2p orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10^-2 nm^-1. A)656.3 nm B)486.2 nm C)364.6 nm D)2.057 × 10^-3 nm

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Quizplus · Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy difference between the 3d and 2p orbitals in hydrogen is given by the Rydberg formula: E = -13.6 eV (1/n1^2 - 1/n2^2), where n1 and n2 are the principal quantum numbers of the orbitals. Plugging in the values for n1 = 3 and n2 = 2, we get E = -1.51 eV. Converting this to joules, we get E = -2.42 × 10^-19 J. Using the equation E = hc/λ, we can solve for λ: λ = hc/E = (6.626 × 10^-34 J s)(3 × 10^8 m/s) / (-2.42 × 10^-19 J) = 656.3 nm.

For Hydrogen,what Is the Wavelength of the Photon Emitted When