0
$\begingroup$

The photoelectric effect is explained as that if a photon of energy more than the work function is absorbed by an electron, the electron will overcome the energy binding material, it is well supported by Bohr's model of atom where an electron can jump to n=$\infty$ provided a photon of required energy is absorbed.

(I am equating the photoelectric electron emission to excitation of an electron from its ground state to n=$\infty$ in Bohr model)

Now as Bohr model is obsolete, How does the uncertainty principle, Hund's rules and probability distribution(which says that an electron ha a non-zero probability of being located at very large distances) deviate or modify or accommodates in photoelectric effect?

Or to put in another way, What is the quantum mechanical explanation of photoelectric effect?

(I would love a simple explanation,understandable to high schooler)

$\endgroup$
  • $\begingroup$ The first paragraph of your question sums up as your answer, provided your definition of quantum mechanics is not different. $\endgroup$ – SchrodingersCat Feb 19 '17 at 11:15
0
$\begingroup$

$$ H \psi = E \psi$$

The energy of a state is given by the quantum mechanical observable, the Hamiltonian. By solving for this equation for a particular potential we can find these eigenstates and eigenvalues. (for a simple system with one valence electron like Rubidium you make the approximation that dynamics can be modeled almost exclusively by the one valence electron, since other the protons and electrons cancel each other out.)

We see that after solving these equations, there's a discrete number of solutions. The eigenvalues associated with these eigenstate/eigenfunctions are the energies.

The process of "absorbing a photon" is simply that a system with fixed eigenstates can only jump from one energy eigenstate to another energy eigenstate. They can't go half-way there, they must go all the way.

There's some more technical complications when you do a full quantum mechanical treatment. Like the quantum mechanical is a composite system so you need to consider the energy of all two objects, and that the energy of the second object will often modify the Hamiltonian significantly. Additionally the term "photon" is also another can of worms, as the real meaning of it typically refers to the discrete units of energy the E-field can take on. But the rough idea is still there.

$\endgroup$
  • $\begingroup$ "One of the primary axioms in quantum mechanics is that the only things you can observe are eigenstates of the system." Is this your personal opinion? Could you give a reference? "Eigenstates of the system" does not even make sense until you specify the observable. If, however, you mean eigenstates of energy, it still is not obvious that such statement is correct. $\endgroup$ – akhmeteli Feb 19 '17 at 10:20
  • $\begingroup$ I realize my language is a little sloppy and fixed it. $\endgroup$ – Steven Sagona Feb 19 '17 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy