0
$\begingroup$

My question is a duplicate of this.
Please consider the equation
$\nabla^2\psi + (2m/\hbar^2)[E-V]\psi=0$ (1)

Potential of electron revolving hydrogen atom is given as

$V=\frac{-e}{4\pi\epsilon_0r}$ (2)

Equation (1) is a second order differential equation in variable $\psi$. $V$ is a continuous variable and takes all values from -$\propto$ to 0 at different distances. What is intuitive understanding for E and hence $\psi$ being discrete?

Consider the thought process, an electron gets closer to nucleus and it's potential rises then it's kinetic energy has to automatically adjust to one of the chosen discrete energy levels

If I draw a comparison with a similar orbital system, Total energy of a satellite revolving earth is given by $E=-\frac{Gm_1m_2}{2r}$
And it is continuous.
I understand that Schrodinger equation is a wave equation and from linked question I take that confined strings can have quantized number of hops. However an electron in an isolated atom is of course excited at one end by potential supplied by nucleus but what confines electron at other end?

I'd like to draw your attention to another similar analogy. Please consider wave equation of EM wave
$\nabla^2E=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2} $ Also has a solution with continuous angular momentum and energy.

We know quantized energy levels are proportional to $\frac{1}{n^2}$ and summation of this series is finite. Is this a reason for energy levels being discrete? (Atom cannot supply infinite energy) If this is so why this doesn't apply to gravitational system?

Please help me with intuitive understanding rather mathematical equations

$\endgroup$
2
$\begingroup$

What confines the electron “at the other end”, i.e., far from the nucleus, is the requirement that the wavefunction be normalizable... in other words, that the total probability of finding the electron somewhere be 1. For this to happen, the wavefunction must go to zero far from the nucleus; otherwise the probability would be infinite. So this is like clamping down one end of a string.

$\endgroup$
1
$\begingroup$

The differential equations and postulates of quantum mechanics were chosen in order to describe the following which cannot be described with classical electromagnetism and orbital equations.

  1. the existence of atoms

  2. spectral lines

  3. photoelectric effect

  4. black body radiation

It is not possible to create a stable solution of classical electromagnetism of an electron around an atom. As rotational trajectories introduce accelerations, the electron would eventually radiate continuous em radiation and spiral on the positive nucleus and neutralize it, no discrete spectra would appear. I have read that metastable solutions may be found ( cannot get my hands on the link at present) but metastable means that the smallest perturbation will start the electrons spiraling down.

The spectra(2) are discrete and characteristic of the atoms in the periodic table (1). This lead to the phenomenological Bohr model, where constraints on the angular momentum by axiom forced discrete energy levels and transitions between them explained (2)

The photoelectric effect showed quantization of energy transferred and could be explained by the Bohr model.(3)

Black body radiation classically had the ultraviolet catastrophe problem, which was solved by the quantization conditions.(4)

The Schrodinger equation with the quantum mechanics postulates turned phenomenology into a consistent mathematical theory which describes innumerable situations at the level where the Planck constant can be considered effectively zero.

Please note that the quantum locations of the electrons are not trajectories , but probability distributions for being at an (x,y,z,t) , called orbitals (not orbits). In quantum mechanics all predictions are on probability distributions, and it works, fitting all data.

$\endgroup$
  • $\begingroup$ Thanks a lot for answer containing keypoints of quantum theory and informative links $\endgroup$ – Zaid Syed M Md Apr 30 at 16:25

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.