What is the wavelength of the photon emitted when an electron in a hydrogen atom which is in the initial state n = 4 jumps to the final state n = 2? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)243 nm B)486 nm C)556 nm D)312 nm E)609 nm

Question

Quizplus · Accepted Answer

The wavelength of the photon emitted can be calculated using the Rydberg formula for the energy difference between two hydrogen atom levels, which is given by $\Delta E = \frac{hcR_H}{n_f^2} - \frac{hcR_H}{n_i^2}$, where $R_H$ is the Rydberg constant for hydrogen ($1.097 \times 10^7 m^{-1}$), $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively, $h$ is Planck's constant, and $c$ is the speed of light. The energy difference $\Delta E$ can then be used to find the wavelength $\lambda$ using the relation $\lambda = \frac{hc}{\Delta E}$. For a transition from $n = 4$ to $n = 2$, the calculation yields a wavelength of approximately 486 nm.

What Is the Wavelength of the Photon Emitted When an Electron