A hydrogen atom emits a photon of wavelength 656 nm. From what energy orbit to what lower - ener

A hydrogen atom emits a photon of wavelength 656 nm. From...

Key Concepts

Spectral Series in Hydrogen

In hydrogen, electron transitions between specific energy levels produce series of spectral lines, such as the Balmer series, which is associated with transitions that end at the second energy level. These series are important in both theoretical calculations and practical identifications of atomic transitions.

Hydrogen Atomic Energy Levels

Hydrogen atoms have discrete energy levels, which means the electron within a hydrogen atom can only exist in specific quantum states characterized by principal quantum numbers. When an electron transitions between these levels, it either absorbs or emits a photon with energy equal to the difference between the levels.

Quantum Electron Transitions

When an electron jumps from a higher energy orbit to a lower energy orbit, the energy difference is released in the form of a photon. This process is governed by quantum mechanics and is fundamental to understanding the emission and absorption spectra of atoms.

Photon Emission and the Energy-Wavelength Relationship

The energy of the emitted photon is directly related to its wavelength by the equation E = hc/?, where h is Planck's constant and c is the speed of light. This relationship ties the observed wavelength of light to the specific energy transition occurring within the atom.

Transcript

00:01 In this exercise, we have a hydrogen atom that has some internal transition, and because of the transition, it emits a photon with a wavelength equals to 656 nanometers.

00:16 And we want to know from what initial energy level and y to what final energy level and f the atom transition.

00:29 So in order to calculate that, the first thing i'm going to do is to calculate the energy of the photon.

00:37 So the energy of the photon is hc over lambda.

00:40 Hc is 12040 electron volts nanometers.

00:45 And lambda is 656 nanometers.

00:50 So the energy is 1 .89 electron volts.

00:55 Okay, that's the energy of a single photon.

00:59 And by conservation of energy, we know that the energy of the photon plus the energy of the final energy level and e and f of the hydrogen atom is equal to the initial energy of the system, which is just the energy of the initial energy level, e and i.

01:24 So we have that the energy of the photon is just e &i minus e and f.

01:32 So what we have to do now is to calculate the energy of some energy levels of the hydrogen atom and see for which n -i and n -f we have that the quantity e -n -i minus e -n -f is equal to the energy of the photon, which is 1 .89 electron volts.

01:56 So that's what i'm going to do.

01:57 I'm going to calculate first the energy of several, at least some energy levels of the hydrogen atom.

02:07 And then i'm going to try to see which ones fit this equation that's highlighted in blue here...

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