Electron transitions

Short Description of Topic

The Main Idea

Electron transition to greater orbits and higher energy levels most commonly occurs when an orbiting electron is struck by a high energy photon. Photons, the quanta of light and all electromagnetic radiation, have no mass but carry energy. This means when a photon collides with an electron, the electron can completely absorb the photon and thus all its energy with no change in mass. The transition of an electron to a higher energy level is only permitted when the electron absorbs energy greater than or equal to the difference between the two energy levels. A photon carrying insufficient energy to trigger a transition will either not be absorbed or will be immediately ejected by the electron, either way the result is no change for the electron. If an electron has excess energy after transitioning to a higher energy level, that is if the electron has more energy than is allowed by the energy formula for its particular level, but not enough energy to transition to a yet higher level, this energy will be ejected in the form of a photon (this is permitted by conservation of mass as photons have no mass and thus their creation results in no net change).

Transition to a lower energy level and smaller radius of orbit or a "downward" transition (corresponding to a decrease in the electron's principle quantum number [math]\displaystyle{ n }[/math]) is in many ways the inverse of transition to a higher energy level and greater radius of orbit or "upward" transition (corresponding to a decrease in the electron's principle quantum number [math]\displaystyle{ n }[/math]). As a result of the first and second laws of thermodynamics, electrons prefer to be in the lowest energy level possible (the smallest value of [math]\displaystyle{ n }[/math]). Thus, upward transition as a result of energy input (i.e. collision with a photon) is rarely long lasting as the electron desires to rid itself of its new energy and transition back to its lowest available energy level and smallest available orbit orbit. In order to achieve this downward transition to an orbit of smaller radius the electron must have the lower energy characteristic of this smaller orbit determined by the energy equation. The excess energy the electron carries which keeps it in its larger orbit cannot disappear due to conservation of energy, so something must be done to shed the undesired energy. In order to achieve this an electron wishing to transition downwards ejects a photon with an energy equal to the exact difference in the electron's energies before and after the transition. Consequently, this energy is also equal to the difference between the allowed energy of the initial and final value of [math]\displaystyle{ n }[/math]. Thus, the photon energy is equivalent to ∆[math]\displaystyle{ E_{n} }[/math].

A Mathematical Model

What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.

A Computational Model

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Examples

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What is the energy of a Hydrogen electron in an orbit of radius .4761 nm? What form of electromagnetic radiation is necessary to free this electron from its orbit?

1) Using the formula for the radius of the orbit, find the value of n for this electron:

[math]\displaystyle{ r = a_{0}n^2 }[/math] where n =1,2,3... and the Bohr Radius [math]\displaystyle{ a_{0} = 0.0529*10^{-9} }[/math]

[math]\displaystyle{ n^2 = \frac{r}{a_{0}} = \frac{.4761*10^{-9}}{0.0529*10^{-9}} = 9 }[/math]

2) Using the value of n, calculate the energy of the electron:

[math]\displaystyle{ E_{n} = \frac{-13.6 eV}{n^2} = \frac{-13.6 eV}{9} = -1.51 eV }[/math]

3) Set the electron's ionization energy (the energy required to free it from its orbit) equal to the energy of a photon ([math]\displaystyle{ E_{photon} = \frac{hc}{λ} }[/math]) and solve for the wavelength λ:

[math]\displaystyle{ E_{Ionization} = |E_{n}| = 1.51 eV }[/math]

[math]\displaystyle{ E_{Ionization} = \frac{hc}{λ} }[/math]

[math]\displaystyle{ λ = \frac{hc}{E_{Ionization}} }[/math]

[math]\displaystyle{ λ = \frac{(4.14*10^{-15} eV*s)(2.998*10^8 m/s)}{1.51 eV} = .822 μm }[/math]

4) Examine the wavelengths of electromagnetic radiation and determine the form with an interval of wavelengths containing .822 μm:

A wavelength of .822 μm is characteristic of ultraviolet light.

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