A hydrogen atom makes a transition from the n = 8 to the n = 3 state.The wavelength of the emitted photon is closest to which one of the following values? (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s) A)9.57 × 10^-7 m B)1.13 × 10^-6 m C)3.12 × 10^-7 m D)4.52 × 10^-6 m E)6.34 × 10^-7 m

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Quizplus · Accepted Answer

The energy difference between two levels in a hydrogen atom is given by the formula: $$E = \frac{hcR}{n_2^2} - \frac{hcR}{n_1^2}$$, where $R$ is the Rydberg constant ($1.097 \times 10^7 m^{-1}$), $h$ is Planck's constant, $c$ is the speed of light, $n_1$ and $n_2$ are the principal quantum numbers of the two levels. The wavelength of the emitted photon can be found from the energy difference using the relation $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength. Plugging in the values for the transition from $n = 8$ to $n = 3$, and solving for $\lambda$, we find that the wavelength is closest to $9.57 \times 10^{-7} m$, which corresponds to choice A.

A Hydrogen Atom Makes a Transition from the N =