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Why is the principal energy of an electron lower for excited electrons in a higher energy state?
The Next CEO of Stack OverflowEnergy 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?
$begingroup$
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
edited Mar 20 at 17:32
andselisk
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
$endgroup$
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
edited Mar 20 at 17:32
andselisk
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
$endgroup$
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
edited Mar 20 at 17:32
andselisk
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
$endgroup$
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
energy electrons
edited Mar 20 at 17:32
andselisk
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
edited Mar 20 at 17:32
andselisk
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
edited Mar 20 at 17:32
andselisk
18.8k660123
edited Mar 20 at 17:32
andselisk
18.8k660123
edited Mar 20 at 17:32
andselisk
18.8k660123
18.8k660123
asked Mar 20 at 16:12
chompionchompion
444
asked Mar 20 at 16:12
chompionchompion
444
asked Mar 20 at 16:12
chompionchompion
444
444
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
$endgroup$
$begingroup$
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
Mar 21 at 4:59
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
$endgroup$
$begingroup$
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
Mar 21 at 4:59
|
show 3 more comments
$begingroup$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
$endgroup$
$begingroup$
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
Mar 21 at 4:59
|
show 3 more comments
$begingroup$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
$endgroup$
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.
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
edited Mar 21 at 6:24
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
answered Mar 20 at 17:32
jheindeljheindel
8,1942553
8,1942553
$begingroup$
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
Mar 21 at 4:59
|
show 3 more comments
$begingroup$
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
Mar 21 at 4:59
$begingroup$
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
Mar 21 at 4:59
$begingroup$
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
Mar 21 at 4:59
|
show 3 more comments
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