The electron in a hydrogen atom,originally in level n = 8,undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm.What is the final level of the electron?
(c=3.00×10⁸ m/s,h=6.63×10^-34 J·s,R_H=2.179×10^-18 J)

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Quizplus · Accepted Answer

The energy of the emitted photon can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the photon. The energy difference between the initial and final levels in a hydrogen atom can also be expressed using the Rydberg formula for the energy levels of hydrogen: $E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $R_H$ is the Rydberg constant for hydrogen, $n_i$ is the initial energy level, and $n_f$ is the final energy level. By equating these two expressions for energy and solving for $n_f$, we find that the final level of the electron is $n_f = 5$.

The Electron in a Hydrogen Atom,originally in Level ​N =

The Electron in a Hydrogen Atom,originally in Level N =