The wavelength of the emitted photon when a hydrogen atom makes a transition from the n = 2 state to the ground state is closest to which of the following values? (c = 3.0 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)122 nm B)203 nm C)243 nm D)389 nm E)411 nm

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Quizplus · Accepted Answer

The energy difference between the n = 2 state and the ground state (n = 1) in a hydrogen atom can be calculated using the formula for the energy levels of a hydrogen atom, $E_n = -\frac{13.6 \text{ eV}}{n^2}$. The energy difference ($\Delta E$) between the n = 2 level and the n = 1 level is then converted to wavelength ($\lambda$) using the formula $\lambda = \frac{hc}{\Delta E}$, where $h$ is Planck's constant and $c$ is the speed of light. The calculated wavelength falls closest to 122 nm, which corresponds to the transition of an electron from the n = 2 state to the n = 1 state in a hydrogen atom.

The Wavelength of the Emitted Photon When a Hydrogen Atom