# Tag Info

2

I will put in my two cents: The Bohr model as such can be saved by postulating standing waves for the electrons. The contrast with the Schrodinger formalism lies not only, as observed by others , that the solutions of the Schrodinger equations are more accurate and can be generalized to complicated potentials, but that the Bohr model is only one step higher ...

1

Here is my answer to this very difficult issue. In my opinion the elementary Schroedinger approach does not solve the problem of radiation. The electron still radiates when it changes its energy level and this process is not described in the elementary Schroedinger model based on the Coulomb potential only. Experiments prove that every level is not stable, ...

2

Have a look at the binding energy per nucleon curve: There are many stable configurations below iron, so the binding energy is not the only criterion for stability. Graph of nuclides (isotopes) by type of decay. Orange and blue nuclides are unstable, with the black squares between these regions representing stable nuclides. The unbroken line passing ...

0

Muonic atoms should be stable in electron-degenerate matter (white dwarf material) as long as the Fermi energy is more than $m_\mu - m_e$. This is more or less exactly a analogy with neutron stability in the nucleus where the the protons are effectively in a degenerate state. Any answer has to forbid electrons (which isn't going to be possible as they share ...

-1

Resolving (Decreasing) energy gradients to lower realized potential states is what drives every process on many levels including evolution. www.intothecool.com Here is a super cool paper that shows the process across universal, biological and socio-economic domains. http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=185965 if ...

3

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...

0

A system that is in thermal contact with it's environment will tend towards both a lower energy state and a higher entropy state. Basically, the energy of the system + environment is fixed, but energy will flow between the two until they are in a state of maximum entropy. It might be more informative to ask why systems tend towards increased entropy. What ...

0

Well, chemical reactions almost always require heat (energy) to take a place, and almost always release heat upon reaction, so by that logic state when elements is unable to keep reacting is a state with insufficient energy or, in other words, lowest energy state (or we probably should say "lower energy state" then one that required for reactions)

0

I'll try to explain with the help of an classical example. Take the situations in the picture above. What you're interested in are the first to cases. The unstable state of equilibrium is such a state that when you slightly displace the ball, it departs from the original position. Being at the top of the hill, it has an excess of potential energy (may it ...

1

Roughly: Becouse $F=-\overrightarrow\nabla U$, with $U$ some potential energy (coud be an effective potential energy). Then, if you aren't in a minimum of potetial, your system isn't in equilibrium. Edit: Can you see that 1. and 2. are stables equilibrium?. I chemistry your effective potential energy is some function called Gibbs free energy.

Top 50 recent answers are included