Question #74da7

1 Answer
Oct 25, 2017

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^2E=mc2

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

p=mvp=mv

Or, solving for mass;

m=p/vm=pv

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

E = p/c c^2E=pcc2

E = pcE=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/pλ=hp

Where hh is Planck's constant., 4.136× 10^-15 "eV" * "s"4.136×1015eVs. Rearranging in terms of pp;

p=h/lamdap=hλ

Plug this momentum term into our energy equation to get;

E = (hc)/lamdaE=hcλ

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")