Mrthdrb

Why do Radio Buttons not fill the entire outer circle?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Why does this expression simplify as such?

Can I say "fingers" when referring to toes?

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

Is it allowed to activate the ability of multiple planeswalkers in a single turn?

Is it necessary to use pronouns with the verb "essere"?

What is Cash Advance APR?

Does the Linux kernel need a file system to run?

How can I write humor as character trait?

Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?

Why Shazam when there is already Superman?

How do I tell my boss that I'm quitting soon, especially given that a colleague just left this week

What is the English pronunciation of "pain au chocolat"?

Will number of steps recorded on FitBit/any fitness tracker add up distance in PokemonGo?

How to make money from a browser who sees 5 seconds into the future of any web page?

Why do ¬, ∀ and ∃ have the same precedence?

Why does AES have exactly 10 rounds for a 128-bit key, 12 for 192 bits and 14 for a 256-bit key size?

What fields between the rationals and the reals allow a good notion of 2D distance?

How many arrows is an archer expected to fire by the end of the Tyranny of Dragons pair of adventures?

Does Doodling or Improvising on the Piano Have Any Benefits?

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

Change the color of a single dot in `ddot` symbol

Why is the principal energy of an electron lower for excited electrons in a higher energy state?

Energy required to remove an electron from HeCan a battery powered by iron and air really provide a feasible power source for automobiles?Calculating the ionization energy for hydrogenLower an electron energy level by pressureIf d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block?Wavelength of an electron removed from an atom of hydrogenWhy are higher-energy bonds preferred over lower-energy ones?State symbol for electronIonisation energy is lower for higher energy shell?What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state?

8

$begingroup$

Several places state the 'principal energy of an electron' can be calculated as such:

$$E = frac2π^2mZ^2e^4n^2h^2$$

Another equation I found was:

$$E = -fracE_0n^2,$$

where $$E_0 = pu13.6 eV~(pu1 eV = pu1.602e-19 J)$$

As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.

However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.

I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.

energy electrons

share|improve this question

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

8

$begingroup$

Several places state the 'principal energy of an electron' can be calculated as such:

$$E = frac2π^2mZ^2e^4n^2h^2$$

Another equation I found was:

$$E = -fracE_0n^2,$$

where $$E_0 = pu13.6 eV~(pu1 eV = pu1.602e-19 J)$$

As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.

However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.

I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.

energy electrons

share|improve this question

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Several places state the 'principal energy of an electron' can be calculated as such:

$$E = frac2π^2mZ^2e^4n^2h^2$$

Another equation I found was:

$$E = -fracE_0n^2,$$

where $$E_0 = pu13.6 eV~(pu1 eV = pu1.602e-19 J)$$

As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.

However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.

I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.

energy electrons

share|improve this question

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

andselisk

18k656119

edited yesterday

andselisk

18k656119

asked yesterday

chompionchompion

444

New contributor

chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

chompionchompion

444

1 Answer
1

active

oldest

votes

9

$begingroup$

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~mathrmeV
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $nrightarrowinfty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.

share|improve this answer

edited 21 hours ago

answered yesterday

jheindeljheindel

8,1592453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My first exposure to this equation was the first, and read the relation between energy and quantum number was inverse. I don't quite see where the minus would come in the 1st equation, but do see it in others. Thank you
  $endgroup$
  – chompion
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
  $endgroup$
  – Ruslan
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's a similar thing for gravitational potential energy. If we set the GPE at infinity to zero, then the GPE for finite $r$ is proportional to $frac -1 r$
  $endgroup$
  – Acccumulation
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ruslan currently it says the energy of an electron in a hydrogen atom is $-E_0$ which means the energy required to ionize the hydrogen atom is $+E_0$.
  $endgroup$
  – jheindel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Citing your first sentence: "we have, $E=-E_0$ which is the energy required to remove an electron...". So, $E$ is the energy required, and, since $E_0>0$, this energy required is $E<0$.
  $endgroup$
  – Ruslan
  23 hours ago
  

  
  
  
  

 | 
show 1 more comment

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "431"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

chompion is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111269%2fwhy-is-the-principal-energy-of-an-electron-lower-for-excited-electrons-in-a-high%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

9

$begingroup$

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~mathrmeV
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $nrightarrowinfty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.

share|improve this answer

edited 21 hours ago

answered yesterday

jheindeljheindel

8,1592453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My first exposure to this equation was the first, and read the relation between energy and quantum number was inverse. I don't quite see where the minus would come in the 1st equation, but do see it in others. Thank you
  $endgroup$
  – chompion
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
  $endgroup$
  – Ruslan
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's a similar thing for gravitational potential energy. If we set the GPE at infinity to zero, then the GPE for finite $r$ is proportional to $frac -1 r$
  $endgroup$
  – Acccumulation
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ruslan currently it says the energy of an electron in a hydrogen atom is $-E_0$ which means the energy required to ionize the hydrogen atom is $+E_0$.
  $endgroup$
  – jheindel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Citing your first sentence: "we have, $E=-E_0$ which is the energy required to remove an electron...". So, $E$ is the energy required, and, since $E_0>0$, this energy required is $E<0$.
  $endgroup$
  – Ruslan
  23 hours ago
  

  
  
  
  

 | 
show 1 more comment

9

$begingroup$

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~mathrmeV
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $nrightarrowinfty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.

share|improve this answer

edited 21 hours ago

answered yesterday

jheindeljheindel

8,1592453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My first exposure to this equation was the first, and read the relation between energy and quantum number was inverse. I don't quite see where the minus would come in the 1st equation, but do see it in others. Thank you
  $endgroup$
  – chompion
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
  $endgroup$
  – Ruslan
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's a similar thing for gravitational potential energy. If we set the GPE at infinity to zero, then the GPE for finite $r$ is proportional to $frac -1 r$
  $endgroup$
  – Acccumulation
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ruslan currently it says the energy of an electron in a hydrogen atom is $-E_0$ which means the energy required to ionize the hydrogen atom is $+E_0$.
  $endgroup$
  – jheindel
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Citing your first sentence: "we have, $E=-E_0$ which is the energy required to remove an electron...". So, $E$ is the energy required, and, since $E_0>0$, this energy required is $E<0$.
  $endgroup$
  – Ruslan
  23 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~mathrmeV
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $nrightarrowinfty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.

share|improve this answer

edited 21 hours ago

answered yesterday

jheindeljheindel

8,1592453

$endgroup$

share|improve this answer

edited 21 hours ago

answered yesterday

jheindeljheindel

8,1592453

answered yesterday

jheindeljheindel

8,1592453

answered yesterday

jheindeljheindel

8,1592453