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I was studying about mechanical energy and I found a line in my book saying that negative value of mechanical energy indicates a bound state. The examples given were electron in an atom, satellite revolving around a planet both are in bound state having negative mechanical energy.

Since I wasn't aware of bound state I searched it on google where I found that "In quantum physics, a bound state describes a system where a particle is subject to a potential such that the particle has a tendency to remain localised in one or more regions of space. The potential may be either an external potential, or may be the result of the presence of another particle"**(CREDIT-WIKIPEDIA)**but throughout the article,i never found terms like negative mechanical energy r satellite revolving around earth.

So the Wiki article made it more confusing for me. I'm stuck here. Shall be obliged if someone help me. I have a little knowledge of Quantum mechanics and a bit more of classical mechanics, so if you are writing an answer, please try to keep it simple.

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    $\begingroup$ What's the actual question? $\endgroup$ – lemon Oct 11 '16 at 21:18
  • $\begingroup$ Through the Question,i want to ask the relation of negative megenical energy and bound state. $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 3:48
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Look up the total mass of two free protons and two free neutrons. Then look at the mass of a helium nucleus, which is made up of two neutrons and two protons, but now they are bound together, you should find that He nucleus is different in mass by around -28MeV. You can find all these figures on wikipedia. That minus sign is important, it indicates that to break up the He nucleus (aka an alpha particle), you need to input 28 MeV. This is called the binding energy, and a negative binding energy, as we have here, tells us that the He nucleus is very stable. I know you asked about electrons but I think this is a better example of exactly the same effect bound electrons have as they are attracted to the nucleus. Think about conservation of energy and $E = mc^2$, as regards the mass of free protons versus bound ones, as the binding energy needs to come from somewhere.

Proton mass : 938.3 MeV

Neutron mass: 939.6 MeV

Total mass of 2 protons and 2 neutrons : 3755.8 MeV

Total bound state (He nucleus) is : 3728.4 MeV. So in mass terms, a He nucleus has 27.4 MeV less mass than 4 free nucleon a and this mass is now the binding energy.

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  • $\begingroup$ If we are using mass energy equivalence,then should't be the mass of alpha particle less then the sum of mass of two protons and two neutrons. $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 3:47
  • $\begingroup$ Yes, a He nucleus does have less mass, than 4 separate particles. I have put that in my post. $\endgroup$ – user108787 Oct 12 '16 at 17:06
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Two masses

The answer is largely a matter of convention.

Consider two masses $m$ and $M$ at a distance $r$ away from each other. They exert a force on each other:

$$F=G\frac{mM}{r^2}$$

If $r\to +\infty$ then $F\to 0$. At an infinite distance there is no attractive force any more, the masses are 'free' and the potential energy of the system is $U_{\infty}=0$.

But to move $m$ from a finite difference $r$ to $r=+\infty$ we need to do work $W$, to overcome the attractive force $F$: $$W=U_{\infty}-U_r>0$$ $$\implies 0-U_r=W$$ So: $$U_r<0$$

But in other cases we use a different convention. If the smaller mass $m$ sat on the surface of the larger mass $M$ (e.g. an object on the surface on the Earth) we usually define that as $U=0$.

Moving the object to some (not too large difference) height $h$ costs work:

$$W=mgh$$

The potential energy $U$ at height $h$ is then:

$$U_h=mgh>0$$

If that sounds contradictory then bear in mind that we're rarely interested in absolute levels of energy and usually only in the changes in energy levels to effectuate certain changes in the system.

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  • $\begingroup$ ,sorry for that but i can't get the conclusion.How can i relate Ur<0 and mgh>0 $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 3:51
  • $\begingroup$ @VidyanshuMishra: as I wrote, it's a matter of convention. In the first case we set $U_{\infty}=0$, in the second $U_h=0$. Now when we calculate how much energy (work) we need to do to move $m$ away from $M$ we get the correct result in both cases. That's because we're interested in the change in energy, not the absolute levels. Energy is relative to a state. $\endgroup$ – Gert Oct 12 '16 at 6:48

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