What is the least amount of energy that can be emitted by an...

What is the least amount of energy that can be emitted by an excited electron in a hydrogen atom falling from an excited state directly to the $n=3$ state? What is the quantum number $n$ for the excited state? Humans cannot visually observe the photons emitted in this process. Why not?

Key Concepts

Visible Light Spectrum

Human vision is sensitive only to a narrow band of the electromagnetic spectrum, typically wavelengths between approximately 400 nm and 700 nm. Photons emitted during certain atomic transitions, such as the n = 4 to n = 3 transition in hydrogen, may have wavelengths in the infrared region, making them invisible to the human eye.

Photon Energy and Wavelength

The energy of a photon is related to its wavelength by the equation E = hc/?. Lower energy photons have longer wavelengths and can fall outside the visible spectrum. For the hydrogen transition from n = 4 to n = 3, the emitted photon has comparatively low energy (about 0.66 eV) and consequently a long wavelength.

Hydrogen Atom Energy Levels

In the hydrogen atom, electrons inhabit quantized energy levels determined by the principal quantum number, n. The energy of each level is given by a formula derived from the Bohr model (E_n = -13.6 eV/n²), meaning that as n increases, the energy becomes less negative (i.e., higher). This quantization is fundamental to understanding how energy changes when electrons transition between states.

Electron Transitions and Photon Emission

When an electron moves from a higher energy level to a lower one, the energy difference is released as a photon. The energy of this photon directly corresponds to the difference between the energies of the initial and final states. In the context of the question, the smallest energy emitted in a transition to n = 3 occurs when the electron falls from the next higher level, which is n = 4.

Transcript

Please give Ace some feedback