The longest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is n = 3,is closest to which one of the following values? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)550 nm B)575 nm C)600 nm D)625 nm E)658 nm

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Quizplus · Accepted Answer

The longest wavelength corresponds to the lowest energy transition. For a hydrogen atom transitioning from n=3, the lowest energy transition is to n=2 (n=3 to n=2). The energy of a photon emitted during a transition is given by the Rydberg formula: $$E = \frac{hcR_H}{n_f^2} - \frac{hcR_H}{n_i^2}$$ where $R_H$ is the Rydberg constant for hydrogen ($1.097 \times 10^7 m^{-1}$), $h$ is Planck's constant ($6.626 \times 10^{-34} J \cdot s$), $c$ is the speed of light ($3.00 \times 10^8 m/s$), $n_i$ is the initial energy level (3), and $n_f$ is the final energy level (2). Rearranging the energy equation to solve for wavelength ($\lambda$) gives: $$\lambda = \frac{hc}{E}$$ Substituting the values and solving for $\lambda$ yields a wavelength closest to 658 nm.

The Longest Wavelength of a Photon That Can Be Emitted