0
$\begingroup$

While going through the quantum mechanical model of atom section in my chemistry textbook, the text states:

The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons and {the energy levels} are allowed solutions of the Schrodinger wave equation.

What does 'direct result of wave-like properties of electrons' mean?

|improve this question
$\endgroup$
  • $\begingroup$ Welcome to chemistry.stackexchange! :) If you are confused by the terminology, then try to replace "direct result of" with "consequence of". If we assume wave-like properties for electrons, then the mathematical derivations naturally lead to quantized energy levels for these electrons. $\endgroup$ – Yoda Apr 30 '18 at 18:50
  • $\begingroup$ I would really appreciate if you could give an insight on how wave like properties of electrons ultimately lead to quantized electronic energy levels $\endgroup$ – Harish singh Apr 30 '18 at 19:21
  • 1
    $\begingroup$ The idea is based on a very simple/basic model of an atom (such as Bohr model) which is that the only energy levels possible are those in which an electron (as a wave ) has an integer number of wavelengths in its 'orbital', i.e. the same wave exactly repeats itself in its orbital. $\endgroup$ – porphyrin May 1 '18 at 7:34
  • 1
    $\begingroup$ While waiting for an answer, perhaps try to read about "standing waves". Maybe this question can help you? $\endgroup$ – Yoda May 1 '18 at 16:30
  • 1
    $\begingroup$ That said, the quote is wrong on many levels. QM states that only some states are observable and points out how to find which ones. It so happens that for atomic levels the resulting math problem might be modified to resemble some oscillation problems, and this is the only reason for similarities. $\endgroup$ – permeakra May 1 '18 at 19:36
0
$\begingroup$

To choose a more basic example, let us consider the particle in a box with

\begin{equation} V(x)={\begin{cases}0 & 0\lt x \lt L\\\infty & \text{otherwise}\end{cases}} \end{equation}

To solve this we use the Schrödinger Equation

\begin{equation} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x) \end{equation}

subject to the boundary conditions \begin{equation} \psi(0)=\psi(L)=0. \end{equation}

The origin of these boundary conditions is, that the wave function needs to be a) continuous and and b) zero outside of the potential well, since probability goes to zero for an infinite potential.

As an ansatz for the wave function we take the solution of the free particle, i.e. planes waves: \begin{equation} \psi(x) = \exp\left(-ikx\right) \end{equation} with \begin{equation} k^2=\frac{2mE}{\hbar^2}. \end{equation} What we now have is a periodic wave function, with period $L$ (or an integer multiple of that) due to the boundary condition, hence \begin{equation} kL = 2\pi n. \end{equation} At this point we have introduce the integer quantum number $n>0$. With this we can now write down wave function and the corresponding energy. Both depend on $n$ and are therefore quantized.

So the origin of the quantized nature here are the boundary conditions. The quantum number naturally show up in the solution of the Schrödinger Equation. Obviously this can happen for other system, such as the Hydrogen atom as well.

As a final side note: The Schrödinger equation does not always lead to quantized solutions, it depends on the system. There are for example continuous states when considering the dissociation of molecules.

|improve this answer
$\endgroup$

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.