Total energy of hydrogen at ground state

[songoku]

Homework Statement

What is the total energy of the hydrogen atom at ground state?
a. 13.6 eV
b. m_pc² + m_nc²
c. m_pc² + m_nc² - 13.6 eV
d. m_pc² + m_nc² + 13.6 eV

Homework Equations

E = -13.6 eV / n²

E = mc²

The Attempt at a Solution

The energy of electron at ground state is - 13.6 eV, so the total energy of hydrogen at ground state will be m_pc² - 13.6 eV but not in the option

Hydrogen does not have neutron so I don't think neutron should be included and I am not sure whether unit of mc² is consistent with 13.6 eV

Thanks

 

[Charles Link]

It looks to me like it should read $m_e c^2$ for the energy of the mass of the electron, rather than $m_n c^2$ which would be the energy of the mass of the neutron. Are you sure you typed up the choices accurately?

 

[songoku]

It looks to me like it should read $m_e c^2$ for the energy of the mass of the electron, rather than $m_n c^2$ which would be the energy of the mass of the neutron. Are you sure you typed up the choices accurately?

I did but I got the question from my friend so maybe he copied it wrongly. Let say m_n should be m_e, so the answer should be m_pc² + m_ec² - 13.6 eV?

- 13.6 eV is already energy of electron at ground state in hydrogen atom, should we still take mass of electron into consideration to find the total energy?

Thanks

 

[PeroK]

I did but I got the question from my friend so maybe he copied it wrongly. Let say m_n should be m_e, so the answer should be m_pc² + m_ec² - 13.6 eV?

- 13.6 eV is already energy of electron at ground state in hydrogen atom, should we still take mass of electron into consideration to find the total energy?

Thanks

Yes. Note that the energy of hydogen is less than the energy of a free proton and a free electron. This implies it takes an input of energy to ionise hydrogen; and, of course, that energy is released when a hydrogen atom forms.

 

[neilparker62]

Yes. Note that the energy of hydogen is less than the energy of a free proton and a free electron. This implies it takes an input of energy to break the hydrogen bond; and, of course, that energy is released when a hydrogen atom forms.

I'm not sure this is entirely correct if you are equating the 13.6 ev to energy needed to break the hydrogen bond (dissociation energy). As far as I understand it the 13.6 ev is the ionization energy of atomic hydrogen (energy needed to enable transition of electron from n=1 - ground state - to n=∞). If you have 'atomic hydrogen' then necessarily the H-H bond has already been broken.

https://research.vu.nl/ws/portalfiles/portal/2485538:

All a bit beyond my 'ken' admittedly!

 

[mjc123]

I think he meant the energy to dissociate the electron from the proton, i.e. the ionisation energy. In chemistry we don't call the nucleus-electron interaction in an atom a "bond". The language suggests the H-H bond in dihydrogen, which is not under consideration here. (The phrase "hydrogen bond" has another meaning again.)

 

[PeroK]

I think he meant the energy to dissociate the electron from the proton, i.e. the ionisation energy. In chemistry we don't call the nucleus-electron interaction in an atom a "bond". The language suggests the H-H bond in dihydrogen, which is not under consideration here. (The phrase "hydrogen bond" has another meaning again.)

Yes, I don't claim to know any chemistry. I just knew $13.6eV$ as the "binding" energy in this case. Anyway, I've corrected it to "ionisation" energy.

 

[neilparker62]

Yes, I don't claim to know any chemistry. I just knew $13.6eV$ as the "binding" energy in this case. Anyway, I've corrected it to "ionisation" energy.

$$ E_i=E_∞-E_1=μc^2\left(1-\sqrt{1-α^2}\right) J $$ with corrections for recoil energy and 'ground state Lamb shift' as per attachment above. It's negative because for some reason the electron is deemed to have zero energy when it's in the excited state (??) and therefore loses energy when it falls back down to ground state.

μ is the reduced electron mass and α the fine structure constant.

 

[PeroK]

$$ E_i=E_∞-E_1=μc^2\left(1-\sqrt{1-α^2}\right) J $$ with corrections for recoil energy and 'ground state Lamb shift' as per attachment above. It's negative because for some reason the electron is deemed to have zero energy when it's in the excited state (??) and therefore loses energy when it falls back down to ground state.

μ is the reduced electron mass and α the fine structure constant.

This is introductory homework. Setting the PE (gravitational or electric) to 0 at "infinity" is fairly standard.

 

[neilparker62]

Except that the energy is supposed to be the sum of potential and kinetic energy and my Physics textbook says at n = ∞ the electron is 'at rest'. I have a conceptual problem with a highly excited electron being 'at rest'. At rest relative to what ? Does it just sit in some kind of suspended motionless state ?

 

[PeroK]

Except that the energy is supposed to be the sum of potential and kinetic energy and my Physics textbook says at n = ∞ the electron is 'at rest'. I have a conceptual problem with a highly excited electron being 'at rest'. At rest relative to what ? Does it just sit in some kind of suspended motionless state ?

If we are looking at hydrogen energy states in the QM model, then:

In the ground state, the expected value of the potential energy is $-2E_1$ and the expected value of kinetic energy is $E_1$, giving the ground state energy $-E_1$.

In higher states, the expected value of the KE decreases, tending to zero, as the potential energy increases. "Excited" doesn't mean "moving fast".

These measurements are assumed in the rest frame of the nucleus (proton).

This is analogous to gravitational orbits, where the kinetic energy reduces with the radius of the orbit.

 

[songoku]

Thank you very much