quantum mechanics - Probability of photon emission

If a photon of a given wavelength is absorbed by an electron (for simplicity, let's assume the electron has only one excited state), does the probability that the electron jumps to its excited state and emits a photon as it falls back depend on the incoming photon's wavelength (is the electron more likely to emit a photon if the incoming photon has an energy much greater than the excitation energy relative to an incoming photon with an energy very close to the excitation energy)?

Answer

There is an equation that helps a lot understanding this issue: Fermi's golden rule

$$ W_{i\rightarrow f}=\frac{2\pi}{\hbar} \left|\left \right|^2 \rho $$

It describes the transition rates from one state to another. $\rho$ is the so called Density of States (DOS) of final states. This system has only two states: The initial state $\left|i\right>$ and the final state $\left|f\right>$.

Therefore the DOS of final states is basically something like this the following, where $\omega_{ph}$ is the photon frequency and $\omega_f=\frac{E_f}{\hbar}$. $$ \rho=\frac{1}{V}\delta(\omega_{ph}-\omega_f) $$

Hence Fermi's Golden Rule for this case reads $$ W_{i\rightarrow f}=\frac{2\pi}{\hbar V} \left|\left \right|^2 \delta(\omega_{ph}-\omega_f). $$ The properties of the Delta-distribution indicate that only photons with the exact right amount of energy can excite the electron to the final state. In reality there are so called "line broadening" mechanisms (e.g. spectral line broadening) that also allow photons with an energy close to the excitation energy to excite the electon.

Conclusion:

Only photons with a wavelength that correlates to an energy very close to the excitation energy of the electron can excite the electron. Photons with less energy won't be absorbed as well as photons with more energy.

I hope this helped you at least a little bit!