What is E in schroedinger equation Is it energy of whole atom or energy of electron only?

Energy of the system. Since the proton is much more massive, to first order it's the energy of the electron, but the use of reduced mass is there to account for that.

What is E in schroedinger equation Is it energy of whole atom or energy of electron only?

 

 

As I understand it the Schroedinger equation is fundamental to quantum mechanics and contains the svi symbol , which is the wave function. ( as like a sine wave ) except it modulates as a probability function as illustrated below.

 

If E is energy then the Schroedinger equations will produce peaks and troughs of probability. Which if you square them they are always Positive .

 

This then shows the position or rather energy levels which are present in the atom. Some Possible , some not possible.

 

 

Wickapedia Sch...Equation

 

 

 

The following sketch ( quickly drawn not a wickpedia drawing ) shows when you turn the Sch...Equat.... into a probability wave by squaring the amplitude you get these peaks and troughs for permitted energy band or, forbidden energy band.

 

Thus No electrons permitted at the center where the proton is. but Some (2) electrons permitted at the first peak( energy band ), and some more electrons say 6 with carbon and 8 with oxygen (max 8 ) in the second peak (energy band )

 

 

Edited March 22, 201312 yr by Mike Smith Cosmos

The energy term in the Hydrogen atom is usually expressed as the sum of the kinetic energy of center-of-mass motion (which can pretty much be attributed to the proton), the remaining kinetic energy (which can be attributed to the electron, as Swansont sais), and the potential energy. The latter is a collective property of the Proton-Electron system, and cannot be sensibly attributed to either of the individual components, in my opinion.

Note that the center-of-mass motion if often ommitted, because it's solutions and the process how they combine into the total solution are assumed to be well-known to the reader (which in my experience usually isn't the case).

In he following discussion please point out where i have gone wrong

final state energy- initial state- energy-=energy change =loss of potential energy=V

E-V=0

There are several things in your statement that could be considered wrong. Most importantly, you didn't express yourself clearly and unambiguously: What system are you talking about? What process? What are V and E, respectively? Is it possible to formulate your pseudo-equation in proper sentences?

hydrogen atom is formed by the combination of proton and electron initially separated by in finite distance hence initial energy is zero E is the energy of hydrogen atom formed after combination of proton and electron hence it is final state energy energetics is

used to get get E-V=0 I think it is clear now

V is the loss in electrostatic potential energy as electron moves towards proton i am sorry i could connect it with my previous post

The electron-proton system contains a kinetic energy term that I expect to be non-zero ("the electron orbits the core"). Hence, I do not think that |E|=|V| is correct.

No, it's not. The system must release energy to become bound, so it will not have the starting energy of the free configuration.

The energy to bind atoms together is known as the Binding Energy. Helium requires less Binding energy than two hydrogen atoms. Thus when hydrogen combines through the root of deuterium to helium , there is a surplus of energy , the net binding energy is less. So energy is given off ( like the Sun ). or a Hydrogen Bomb.

 

 

 

 

 

Sorry the line is a bit of a wobble. ( must be the cold )

Edited March 24, 201312 yr by Mike Smith Cosmos

Is it correct to say that half of the electrostatic potential energy liberated as the electron moves towards proton to form hydrogen atom is radiated and the other half is used as kinetic energy of electron but it is contrary to quantum mechanics which denies orbital motion to electron Please clarify it

Is it correct to say that half of the electrostatic potential energy liberated as the electron moves towards proton to form hydrogen atom is radiated and the other half is used as kinetic energy of electron but it is contrary to quantum mechanics which denies orbital motion to electron Please clarify it

 

That is not contrary. QM applied to the H atom does not have any trajectories, but there is still the same relationship with the kinetic and potential energies.

in my opinion E is neither the energy of hydrogen atom nor that of electron It is the energy radiated as the electron starting from infinity occupies the specified slot in hydrogen atom

in my opinion E is neither the energy of hydrogen atom nor that of electron It is the energy radiated as the electron starting from infinity occupies the specified slot in hydrogen atom

 

The energy of the atom (which is the binding or ionization energy) and the energy emitted to form it are equal, so it's both.

if E is binding energy ,what is( E-V)

E-V will be the kinetic energy, assuming the use of terminology is consistent.

 

Where are you getting E-V = 0?

 

The basic setup of the Schrödinger equation is that the total energy (E) is the sum of the kinetic energy and potential energy