Why is the principal energy of an electron lower for excited electrons in a higher energy state?
$begingroup$
Several places state the 'principal energy of an electron' can be calculated as such:
$$E = frac{2π^2mZ^2e^4}{n^2h^2}$$
Another equation I found was:
$$E = -frac{E_0}{n^2},$$
where $$E_0 = pu{13.6 eV}~(pu{1 eV} = pu{1.602e-19 J})$$
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
edited Mar 20 at 17:32
andselisk
18.4k656121
asked Mar 20 at 16:12
chompionchompion
444
New contributor
chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Several places state the 'principal energy of an electron' can be calculated as such:
$$E = frac{2π^2mZ^2e^4}{n^2h^2}$$
Another equation I found was:
$$E = -frac{E_0}{n^2},$$
where $$E_0 = pu{13.6 eV}~(pu{1 eV} = pu{1.602e-19 J})$$
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
edited Mar 20 at 17:32
andselisk
18.4k656121
asked Mar 20 at 16:12
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:
$$E = frac{2π^2mZ^2e^4}{n^2h^2}$$
Another equation I found was:
$$E = -frac{E_0}{n^2},$$
where $$E_0 = pu{13.6 eV}~(pu{1 eV} = pu{1.602e-19 J})$$
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
edited Mar 20 at 17:32
andselisk
18.4k656121
asked Mar 20 at 16:12
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.
$endgroup$
Several places state the 'principal energy of an electron' can be calculated as such:
$$E = frac{2π^2mZ^2e^4}{n^2h^2}$$
Another equation I found was:
$$E = -frac{E_0}{n^2},$$
where $$E_0 = pu{13.6 eV}~(pu{1 eV} = pu{1.602e-19 J})$$
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
energy electrons
edited Mar 20 at 17:32
andselisk
18.4k656121
asked Mar 20 at 16:12
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 Mar 20 at 17:32
andselisk
18.4k656121
asked Mar 20 at 16:12
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 Mar 20 at 17:32
andselisk
18.4k656121
edited Mar 20 at 17:32
andselisk
18.4k656121
edited Mar 20 at 17:32
andselisk
18.4k656121
18.4k656121
asked Mar 20 at 16:12
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 Mar 20 at 16:12
chompionchompion
444
asked Mar 20 at 16:12
chompionchompion
444
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.
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.
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add a comment |
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1 Answer
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$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrm{eV}
$$
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,
$$
E=-E_0/4=-3.4~mathrm{eV}
$$
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.
answered Mar 20 at 17:32
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
|
show 1 more comment
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1 Answer
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$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrm{eV}
$$
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,
$$
E=-E_0/4=-3.4~mathrm{eV}
$$
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.
answered Mar 20 at 17:32
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
|
show 1 more comment
$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrm{eV}
$$
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,
$$
E=-E_0/4=-3.4~mathrm{eV}
$$
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.
answered Mar 20 at 17:32
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
|
show 1 more comment
$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrm{eV}
$$
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,
$$
E=-E_0/4=-3.4~mathrm{eV}
$$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1592453
$endgroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrm{eV}
$$
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,
$$
E=-E_0/4=-3.4~mathrm{eV}
$$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1592453
edited 2 days ago
answered Mar 20 at 17:32
jheindeljheindel
8,1592453
answered Mar 20 at 17:32
jheindeljheindel
8,1592453
answered Mar 20 at 17:32
jheindeljheindel
8,1592453
8,1592453
$begingroup$
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
|
show 1 more comment
$begingroup$
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
$begingroup$
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
Mar 20 at 17:50
$begingroup$
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
Mar 20 at 17:50
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
Mar 20 at 21:47
$begingroup$
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
Mar 20 at 22:26
$begingroup$
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
Mar 20 at 22:26
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
@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
Mar 21 at 1:20
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
|
show 1 more comment
chompion is a new contributor. Be nice, and check out our Code of Conduct.
chompion is a new contributor. Be nice, and check out our Code of Conduct.
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