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].

The negative energies of the electrons are what keeps them in orbit around the nucleus, but it is possible to free the electrons from this orbit by bringing this energy to 0 or any positive energy. Because the electron's energy while in orbit around the nucleus is negative a positive energy of equal magnitude is required to free it from its orbit. Hence why the electron's energy found using [math]\displaystyle{ E_{n} }[/math] is referred to as its binding energy, because it represents the hurdle that must be surmounted in order to remove the electron from orbit.

The increasing ease with which an electron may be freed as it transitions to larger orbits and thus higher energy levels is an interesting and intuitive consequence of this phenomenon.

Mathematical Model

where [math]\displaystyle{ n }[/math] = 1,2,3...

• Radius of orbit: [math]\displaystyle{ r_{n} = a_{0}n^2 }[/math]
  • Where [math]\displaystyle{ a_{0} = {\frac{4π ε_{0}ħ^2}{m_{e}e^2}} = 0.0529 }[/math] nm
• Electron energy level or Binding Energy: [math]\displaystyle{ E_{n} = {\frac{13.6 eV}{n^2}} }[/math]
• Photon frequency: [math]\displaystyle{ λ = \frac{hc}{E_{photon}} }[/math]
  • Where h is Planck's constant.

Derivations found here.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

Simple

Will a photon with energy [math]\displaystyle{ E_{photon} = 10.4eV }[/math] be able to transition a hydrogen electron in the ground state to the n = 2 level?

1. [math]\displaystyle{ E_1 = \frac{-13.6}{1} = -13.6eV }[/math]
   
2. [math]\displaystyle{ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4eV }[/math]
   
3. [math]\displaystyle{ E_2 - E_1 = -3.4eV - -13.6eV = 10.2eV }[/math], so it requires [math]\displaystyle{ 10.2eV }[/math] to transition an electron from the ground state to the second level.
   
4. [math]\displaystyle{ 10.4eV \ge 10.2eV }[/math]
   

Yes, the photon has enough energy to bump the electron from the ground state to the second energy level.

Middling

If an electron orbiting hydrogen starts in the n = 4 orbit and ends in the ground state, how many photons with different energies can the atom emit?

The different possible orbital transitions possible in the atom are (where the numbers indicated are possible values of n):

4 -> 3

4 -> 2

4 -> 1

3 -> 2

3 -> 1

2 -> 1

There are 6 different possible transitions, which correspond to 6 different energy levels the photons emitted from these transitions can have.

Difficult

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:
   1. [math]\displaystyle{ r = a_{0}n^2 }[/math]
   2. where n =1,2,3... and the Bohr Radius [math]\displaystyle{ a_{0} = 0.0529*10^{-9} }[/math]
   3. [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:
   1. [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:
   1. [math]\displaystyle{ E_{Ionization} = |E_{n}| = 1.51 eV }[/math]
   2. [math]\displaystyle{ E_{Ionization} = \frac{hc}{λ} }[/math]
   3. [math]\displaystyle{ λ = \frac{hc}{E_{Ionization}} }[/math]
   4. [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:
   1. A wavelength of .822 μm is characteristic of ultraviolet light.

History

The Franck-Hertz Experiment

See also

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

Further reading

Books, Articles or other print media on this topic

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page