Question

Question #74da7

Answer 1

There are a few relativistic equations that come together to express the relationship between energy and wavelength.

Explanation:

Lets start with Einstein's rest energy equation;

`E = mc^2`

We know that photons are massless, however, they do have momentum. Momentum, `p`, is defined as;

`p=mv`

Or, solving for mass;

`m=p/v`

Now we can replace the mass term in the energy equation. Note that for a photon, `v=c`.

`E = p/c c^2`

`E = pc`

Now we just need to find the momentum of a photon. In a relativistic frame, particles have a wavelength given by the de Broglie wavelength;

`lamda = h/p`

Where `h` is Planck's constant., `4.136× 10^-15 "eV" * "s"`. Rearranging in terms of `p`;

`p=h/lamda`

Plug this momentum term into our energy equation to get;

`E = (hc)/lamda`

When an atom releases a photon, due to electron transition, the energy of that photon will correspond to the change in energy of the atom.

`Delta E_"atom" = (hc)/lambda`

However, in a lab setting, it can be difficult to measure the `Delta E` of a single atom, so we can measure the `Delta E` of a mole of atoms.

`Delta E_"mole" = (N hc) / lamda`

Of course, we can rewrite this to find wavelength in terms of energy.

`lamda = (Nhc)/(Delta E_"mole")`

Here, `N Delta E` would be the total amount of energy released per mole of atoms. `Nhc` are all constants, the product of which is `1.19627 * 10^5 "kJ nm/mole"`.

`lamda = (1.19627 * 10^5 "kJ nm/mole")/(Delta E_"mole")`