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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?
$begingroup$
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 6 hours ago
andselisk
18k656119
asked 7 hours ago
chompionchompion
314
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
edited 6 hours ago
andselisk
18k656119
asked 7 hours ago
chompionchompion
314
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
edited 6 hours ago
andselisk
18k656119
asked 7 hours ago
chompionchompion
314
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:
$$E = frac2π^2mZ^2e^4n^2h^2$$
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 6 hours ago
andselisk
18k656119
asked 7 hours ago
chompionchompion
314
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 6 hours ago
andselisk
18k656119
asked 7 hours ago
chompionchompion
314
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 6 hours ago
andselisk
18k656119
edited 6 hours ago
andselisk
18k656119
edited 6 hours ago
andselisk
18k656119
18k656119
asked 7 hours ago
chompionchompion
314
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 7 hours ago
chompionchompion
314
asked 7 hours ago
chompionchompion
314
314
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.
chompion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is 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 6 hours ago
jheindeljheindel
8,1442453
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour ago
add a comment |
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1 Answer
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active
oldest
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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 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 6 hours ago
jheindeljheindel
8,1442453
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is 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 6 hours ago
jheindeljheindel
8,1442453
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is 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 6 hours ago
jheindeljheindel
8,1442453
$endgroup$
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~mathrmeV
$$
which is 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 6 hours ago
jheindeljheindel
8,1442453
answered 6 hours ago
jheindeljheindel
8,1442453
answered 6 hours ago
jheindeljheindel
8,1442453
answered 6 hours ago
jheindeljheindel
8,1442453
8,1442453
$begingroup$
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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour ago
add a 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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour 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
6 hours 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
6 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a 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.
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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