# Is the energy of the photon coming in greater than or less than the energy of the electronic transition?

Jun 22, 2017

Neither. It has the energy given by the difference in energy, ${E}_{f} - {E}_{i}$.

When photons are absorbed to excite an electron from initial energy ${E}_{i}$ to final energy ${E}_{f}$, they must account for the difference in energy, $\Delta E = {E}_{f} - {E}_{i}$ in order for the electron to transition upwards by that energy.

Suppose you wanted to go from ${E}_{1}$ to ${E}_{2}$ in the hydrogen atom. Then,

${E}_{\text{photon" = DeltaE = -"13.6 eV}} \left(\frac{1}{2} ^ 2 - \frac{1}{1} ^ 2\right)$

would be the energy that the one absorbed photon needs to be in order to get the electron from $n = 1$ (where ${E}_{1} = - \text{13.6 eV}$) to $n = 2$ (where ${E}_{2} = - \text{3.4 eV}$).

If you had $\text{13.6 eV}$ for that photon, you wouldn't be going from ${E}_{1}$ to ${E}_{2}$; you'd be going way past that, since the difference is only $+ \text{10.2 eV}$. And since the difference is that, you can't have $\text{3.4 eV}$ available and bridge that gap.