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Using electrical potential energy $V=\frac{1}{4\pi \varepsilon_0} \frac{Q_1 Q_2}{r}$ , a particle further away from nucleus has lower magnitude of energy.

Using Coulomb's law, a particle further away from nucleus experiences weaker attraction, hence less energy is needed to maintain orbit$^\star$ around that e-shell compared to a electron shell closer to nucleus, hence the one closer to nucleus supposedly should have higher energy.

$^\star$I know in reality $e^-$ does not orbit around a atom, but its position exists as a probability density or radial probability function.

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    $\begingroup$ In a hurry, so I'll just give you a hint: What's the sign of Q1? Q2? What effect does that have on your reasoning that the potential energy decreases as r increases? $\endgroup$ – Brionius May 16 '16 at 10:50
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    $\begingroup$ Moreover, don't consider the absolute magnitudes, it will just cause confusion. You should be considering the difference in PE between two positions... $\endgroup$ – lemon May 16 '16 at 10:53
  • $\begingroup$ Ahh yes ! Okay, i understand better now. Yes, no energy is needed for an electron to maintain an orbit as acceleration is perpendicular to direction of motion Removing e- from a atom or transitioning it to a higher energy level requires energy. I realised my misconception. What about the concept of why electron shells of higher principal quantum number have higher energy level ? $\endgroup$ – De Day May 16 '16 at 12:42
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The potential energy stored in a two like charge system will increase with decrease in distance between them. While for a two unlike charge system, the potential energy decreases with decrease in distance (means potential energy gets liberated if they come close), accounting for increase in attraction.

In the equation, you provided, the potential energy in the nucleus-electron system is negative. This means the potential energy of the system is liberated and hence indicate attraction of the nucleus with the electron (this is how they attain stability).

Hence a system comprising of an electron far off from the nucleus will have high potential energy stored in it, indicating they have sufficient potential energy that can overcome the attractive forces (means the attractive forces between the electron and the nucleus is less). This means the potential energy liberated by an electron far from the nucleus is very less. Hence the outermost electrons are less stable.

For an electron very close to the nucleus, the potential energy is minimum, which means the system comprising of nucleus and a nearer electron liberates most of it's potential energy (so that the system will now have a lesser potential energy) to have an increased attractive force, which in turn corresponds to maximum stability.

So a large amount of energy is required to liberate an electron from an inner most shell rather than an electron from the outermost shell. This is why we say that the electron in the outermost shell has a higher (potential) energy than the inner most shells. So a less amount of energy is needed to liberate the electron from the outermost shell.

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  • $\begingroup$ <For an electron very close to the nucleus, the potential energy is minimum>i could not understand the statement as the potential energy of a particle is the amount of energy stored in it due to the presence of a field i.e. the nuclear charge field and its value is larger when you go closer. $\endgroup$ – drvrm May 16 '16 at 17:05
  • $\begingroup$ The potential energy is negative means the energy is liberated. More attractive force correspond to less potential energy. $\endgroup$ – UKH May 17 '16 at 3:53
  • $\begingroup$ If the electron close to the nucleus is highly energetic it will be unstable. Potential energy is stored in the system. But the energy is negative. As you go far and away from the nucleus more less negative will be the potential energy. Which is greater- Less negative or more negative? $\endgroup$ – UKH May 17 '16 at 3:55
  • $\begingroup$ The potential of the positive charge is reduced by the presence of a negative charge. This is how the potential energy stored in the system is liberated. When two like charges come close work is needed to be done on the system which will get added as the potential energy of the system. In the case of two unlike charges work is done by the system itself which corresponds to a decrease in the potential energy stored in the system. So potential energy will be minimum for two unlike charges while maximum for two like charges $\endgroup$ – UKH May 17 '16 at 4:13
  • $\begingroup$ i think the energy which is binding a particle to a system is best judged by the the energy required by an agency to take the body to an unbound state and that is the crux of the matter- the states closer to a nucleus are such that you need more energy to get the particle free-so its a higher bound state. $\endgroup$ – drvrm May 17 '16 at 5:52
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The energy in a level $n$ is given by $$E = - \frac{Z^2 R_E}{n^2} $$ where $R_E$ is the Rydberg energy ($R_E = 13.6\mathrm{eV}$).

Therefore, greater $n$ means lower energy (in absolute value), i.e., the electron is less bounded.

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  • $\begingroup$ Okay wow, that's true. I agree with you. Thanks for your reply! However there's one aspect i don't understand. In chemistry, the electron shell of higher principal quantum number, has higher energy levels !!! . $\endgroup$ – De Day May 16 '16 at 12:15
  • $\begingroup$ @DeDay both things are true. Higher quantum numbers have higher energy, and higher quantum numbers have lower absolute value of energy. However, "absolute value of energy" is a meaningless quantity, which is why they probably emphasize the fact that higher quantum numbers have higher energy. $\endgroup$ – Jahan Claes May 16 '16 at 12:29
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    $\begingroup$ It would be very nice to explain what all the symbols mean (Z is missing), and what is the meaning of Rydberg's energy. Pretty simple but would probably make the answer substantially more informative with a little of context added. $\endgroup$ – luk32 May 16 '16 at 23:47
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    $\begingroup$ I ageee with luk32. I cant find the above equation from here (en.m.wikipedia.org/wiki/Rydberg_constant) but manage to find a similar eqn. I wonder what is z too. I'm guessing proton number. $\endgroup$ – De Day May 20 '16 at 11:25
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By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level

Or

EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level

Thanks to everyone that helped !

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By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<

Or

EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<

Thanks to everyone that helped !<

I beg to differ on the above explanation provided by the author @ De Day:

The highest energy acquired by an electron is at K shell , and slowly energy decreases as one moves to L,M,N ...shells.

the confirmation is the energy required to take out a K-shell electron is highest and in X-ray emission the high speed cathode electrons knock out K-shell electrons and it needs about 20-25 keV of energy .

Therefore I wish to add that energy levels which are closest to the nucleus is at the highest and the above contention by the author is not correct.

Moreover if a K-shell electron is knocked out and a vacancy is created then any transition from L.M....levels leads to emission lines of lowest wavelength and highest frequency X-rays characteristic lines . This energy packet contains the difference of energy levels of the atom. the magnitude of this energy also suggests that the E(K)-E(m)= h. frequency . is largest.

I think the confusion is that the bound states total energy is sum of its K.E. and P.E. and total energy has to be negative for bound states and it's highest as one moves closer to the nuclear charge +ze.

By just thinking about the potential energy, one has to consider that the nuclear charge field has done work on the electron to get it to a shell radius and this work done is highest if one goes closer.

the test is to supply energy to pull out a K-electron and the value of energy needed to extract will again be largeer than L, M,.. and other shell electrons.

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  • $\begingroup$ If that's the case then why an excited electron go to a higher energy state? For example when excited an electron in the K shell it will go up to the L shell? The experimental justification you provided is true, but the reasoning is poor. $\endgroup$ – UKH May 17 '16 at 7:32
  • $\begingroup$ <why an excited electron go to a higher energy state> $\endgroup$ – drvrm May 17 '16 at 7:39
  • $\begingroup$ Have some read on atomic physics electrodynamics quantum mechanics etc. It will help $\endgroup$ – UKH May 17 '16 at 7:42
  • $\begingroup$ <why an excited electron go to a higher energy state>@unnikrishnan the electron when absorbs energy goes to higher shells , having higher n-values , not higher energy as this movement gets it to closer to continuum. the deepest energy levels are the K- level - you are in a potential well and more deeper you go you will reach more negative energy value.common sense- -10ev is more deeper than -5 ev. as in +ve direction one has higher values similarly in -ve direction one has higher values to -infinity. $\endgroup$ – drvrm May 17 '16 at 7:50
  • $\begingroup$ -10 eV is less than -5 ev $\endgroup$ – UKH May 17 '16 at 8:12

protected by Qmechanic May 16 '16 at 13:50

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