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In the definition of the ground state of an atom, it is given that

The ground state is one in which the electrostatic energy of attraction is minimum. This state is called the ground state of the atom. enter image description here

But, when the atom is in its ground state then the electron and the nucleus will be closest as compared to when they are in the excited state. Isn't it?

Then, the electrostatic force of attraction should be maximum as it is proportional to 1/r^2.

How is it a minimum? What am I missing?

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  • $\begingroup$ This isn't really important, but the electrostatic force and electrostatic energy are different. The force is proportional to $\frac{1}{r^2}$, while the energy is proportional to just $\frac{1}{r}$. $\endgroup$
    – gardenhead
    Aug 24 '20 at 13:50
  • $\begingroup$ More accurately, the energy is proportional to $-1/r$. (Yes, technically it is also proportional to $1/r$, but then the constant of proportionality is negative.) $\endgroup$
    – G. Smith
    Aug 24 '20 at 15:55
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Yes your thinking is correct and the electrostatic force of attraction is maximum in ground state. But that does not change the fact that electrostatic energy is minimum.


Consider this analogy:

Suppose you drop a ball from a height.It will reach the ground - its most stable state. In the ground state,its Potential Energy is minimum and the force of attraction is maximum.


Get it?

You are misinterpreting energy for force here.

Further, in general, any system has a tendency to attain least potential which is its most stable state.


P.S. If you still have any doubt,comment below.

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  • $\begingroup$ Keep it simple. All the OP. missed is that the energy is negative. $\endgroup$
    – my2cts
    Aug 24 '20 at 6:34
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What you are missing is that the electrostatic potential energy between the nucleus and the electrons is negative. “Minimum” here means “as negative as possible”.

For example, for hydrogen the electrostatic PE in the ground state is -27.2 eV. In the first excited state it is -6.8 eV.

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