Bohr atom: A hydrogen atom initially in the n = 6 state decays to the n = 2 state. The emitted photon is detected in a photographic plate. What is the wavelength of the detected photon? The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s)

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Quizplus · Accepted Answer

The energy difference between the initial and final states in the hydrogen atom can be calculated using the formula for the energy levels of a hydrogen atom, $E_n = \frac{-13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. The energy of the emitted photon is the difference in energy between the two states, which can be calculated and then converted to wavelength using the formula $\lambda = \frac{hc}{E}$, where $h$ is Planck's constant, $c$ is the speed of light, and $E$ is the energy of the photon. The calculation will show that the wavelength of the emitted photon corresponds to the option given as 410 nm.

Bohr Atom: a Hydrogen Atom Initially in the N =