Electron Transitions – Learn

Electron levels and the conservation of energy

The Law of Conservation of Energy is one of the most important laws underpinning all of physics, it is crucial in understanding a vast array of principles at the core of this course. Simply, energy cannot be created or destroyed only transformed from one form to another.

To consider how this applies to electron orbitals, the electric field surrounding the nucleus (just a proton in the case of hydrogen) must first be assessed. The diagram below shows the electric field surrounding a point charge, like a proton. Positions P and Q represent different possible locations of an electron. Obviously, an electron situated at either position would experience a force of attraction in accordance with Coulomb’s Law. But which has the greater potential energy?

If the electron started close to the nucleus, more work would need to be done against the electric field to move it to position Q than position P. Hence the energy stored in an electron at position Q would be greater than at P. 

When this principle is applied to Bohr’s model of the atom, it becomes apparent that electrons in each stable orbit have a specific amount of energy attributed to them, as a consequence of their fixed distance from the nucleus. When discussing this, it is important to consider that electric fields around a point charge obey the inverse square law, significantly decreasing in strength as distance increases, similar to gravitational fields around bodies with mass. As a result of this, there are different amounts of energy associated with the transition of an electron to different orbits separated by similar increments. For example, it takes more energy to move an electron from the ground state (n=1) to the first excited state (n=2), than it does from n=2 to n=3. This is often represented in diagrams, like below, by greater distances between orbitals. An understanding of the other series shown in this diagram is not necessary for the HSC Physics course, but it is nice to put the Balmer series into perspective. 

Be careful when reading questions regarding electron orbitals, as the language can sometimes be confusing. The lowest energy state is called the ground state (n=1), however the second energy level (n=2) is sometimes called the first excited state. This is occasionally confused with the ground state, so ensure that questions are read carefully. 

The amount of energy required to move between each level is exactly equivalent to the amount carried by the photon which is absorbed to move up orbits, the emitted when moving down orbits. This is essential to satisfy the conservation of energy.

Deriving an expression for the energy of a photon in terms of wavelength

c=f\lambda

f=\ \cfrac{c}{\lambda}

E=hf

E=\cfrac{hc}{\lambda }

Example 1:

Determine the energy released as an electron falls from the third excited state to the second excited state.

Before commencing, heed the warning above and think carefully about the energy states involved.

Answer:

\cfrac { 1 }{ \lambda } =R\left[ \cfrac { 1 }{ n_{ f }^{ 2 } } -\cfrac { 1 }{ n_{ i }^{ 2 } } \right]

\cfrac { 1 }{ \lambda } =1.097\times 10^{ 7 }\left[ \cfrac { 1 }{ 3^{ 2 } } -\cfrac { 1 }{ 4^{ 2 } } \right]

\cfrac { 1 }{ \lambda } =533263.89

\lambda =1.86\times 10^{ -6 }m

E=\cfrac { hc }{ \lambda }

E=\cfrac { 6.626\times 10^{ -34 }\times 3\times 10^{ 8 } }{ 1.86\times 10^{ -6 } }

E=1.06\times 10^{ -19 }J

E=0.66eV