Ionization Energy: Solving for n and Why I Was Wrong

[jjson775]

Homework Statement

A photon with energy 2.28 eV is barely capable of causing a photoelectric effect on a sodium plate. If the photon instead absorbed by a hydrogen atom, find a) the minimum n for a hydrogen atom that can be ionized by the photon and b) the speed of the released electron.

Relevant Equations

1. En = -13.606/n^2
2. U (r) = -Ke x(e^2/r)
3. rn =n^a0

I got the right answers but have 2 questions. My first attempt was to use Equation 2 and solve for r then use Equation 3 and solve for n. My reasoning was that the photon needed to overcome the potential energy of the atom to release the electron. This gave me n =3.44. Why was I wrong?

Then I reviewed the Bohr atom and saw that Ionization energy to completely release the photoelectron is given by Equation 1 and solved the problem. I got n = 2.44. Since n must be an integer, the correct n must be 3, right?

 

[etotheipi]

I got the right answers but have 2 questions. My first attempt was to use Equation 2 and solve for r then use Equation 3 and solve for n. My reasoning was that the photon needed to overcome the potential energy of the atom to release the electron. This gave me n =3.44. Why was I wrong?

The equation for potential energy should be$$U(r) = \frac{-ke^2}{r}$$But you mustn't forget to account for the kinetic energy that the system already has. The total energy of the hydrogen atom is actually the sum of this potential energy of interaction and the kinetic energy$$E = U(r) + T = -\frac{ke^2}{r} + \frac{ke^2}{2r} = -\frac{ke^2}{2r} = -\frac{13.6 \text{eV}}{n^2}$$Ionisation occurs when a photon can supply $\frac{ke^2}{2r}$ of energy to the atom, such that the final state is of zero energy (zero kinetic + zero potential).

 

[kuruman]

But you mustn't forget to account for the kinetic energy that the system already has.

I do not believe that any kinetic energy is involved here. The CM of the electron and the proton is assumed to be at rest. The $E_n=-\dfrac{13.6~\mathrm{eV}}{n^2}$ is the total energy of the atom relative to a "zero" of energy wherein the proton and the electron are very far apart and at rest. When one adds that much energy to the atom, an electron at rest is promoted from level $n$ to the continuum; when more than that energy is added to the atom, the electron is promoted to the continuum not at rest, but with kinetic energy equal to the "more than that." Note that this treatment ignores momentum conservation and the recoil energy of the atom

 

[etotheipi]

I do not believe that any kinetic energy is involved here. The CM of the electron and the proton is assumed to be at rest. The $E_n=-\dfrac{13.6~\mathrm{eV}}{n^2}$ is the total energy of the atom relative to a "zero" of energy wherein the proton and the electron are very far apart and at rest. When one adds that much energy to the atom, an electron at rest is promoted from level $n$ to the continuum; when more than that energy is added to the atom, the electron is promoted to the continuum not at rest, but with kinetic energy equal to the "more than that." Note that this treatment ignores momentum conservation and the recoil energy of the atom

Virial theorem guarantees that, for a bound state like a hydrogen atom, $\langle E \rangle = -\langle T \rangle = \langle U \rangle /2$, there are details here:

http://muchomas.lassp.cornell.edu/8.04/Lecs/lec_TISE/node8.html

Also, the kinetic energy of the proton/electron in the centre of mass frame is not zero, and the total kinetic energy is the sum of the centre of mass kinetic energy plus that w.r.t. the centre of mass.

The $E_n=-\dfrac{13.6~\mathrm{eV}}{n^2}$ is the total energy of the atom relative to a "zero" of energy wherein the proton and the electron are very far apart and at rest.

The $E_n$ is an eigenvalue of energy, which is a sum of the potential energies and kinetic energies of the system. At infinity, $\langle T \rangle = \langle U \rangle = E = 0$. The potential energy is also defined only up to a constant, so $E_n$ is also defined up to a constant.

 

[kuruman]

The $E_n$ is an eigenvalue of energy, which is a sum of the potential energies and kinetic energies of the system. At infinity, $\langle T \rangle = \langle U \rangle = E = 0$. The potential energy is also defined only up to a constant, so $E_n$ is also defined up to a constant.

Right, the $E_n$ is an eigenvalue of energy of the system. Suppose you are look at this system in which its CM is at rest and you send in a photon of energy $\epsilon.$ What can conceivably happen? The possible answers are something and nothing. Nothing is not interesting so let's look at something that would involve the absorption of the photon. One possibility would be for the system's CM to acquire kinetic energy equal to $\epsilon.$ We can also have internal conversion of energy in which (a) the system's eigenstate just changes from $E_n$ to $E_n'$ (excitation) or (b) the system's eigenstate changes from $E_n$ to $E_{\infty}$ (if that's an eigenstate) and an electron appears with generally non-zero kinetic energy (ionization).

My point is that when you write the energy balance equations for these possibilities, $\langle T \rangle$ and $\langle U \rangle$ do not enter the picture and are irrelevant to answering OP's question. If you add 13.6 eV to a hydrogen atom in the ground state, you get to ionize it, i.e. you end up with a proton and an electron at rest very far from each other. The same result is achieved if you add 3.4 eV to the atom in the first excited state and so on. The energy balance equation is simple with $$E_{\rm{before}}=\epsilon+ \langle T \rangle_{n}+\langle U \rangle_{n} =\epsilon+ E_n~~~{\rm{and}}~~~ E_{\rm{after}}=\langle T \rangle_{\infty}+\langle U \rangle_{\infty}=\frac{1}{2}mv^2$$I am trying to understand your reminder to OP "to account for the kinetic energy that the system already has."

 

[jjson775]

This is going over my head. My question is this: The electron is bound to the proton nucleus by potential energy, Why can't you just consider the energy of the photon and solve for r and n? If the energy (2.28 eV) equals Ke, why can't it knock the electron away, like with the photoelectric effect?

 

[etotheipi]

I am trying to understand your reminder to OP "to account for the kinetic energy that the system already has."

I try to remind that the energy $E_n = \frac{-13.6 \text{eV}}{n^2}$ is the sum of a potential and a kinetic term, since from what I could tell OP had tried to "re-derive it" but had considered only the potential energy of interaction.

 

[kuruman]

Why can't you just consider the energy of the photon and solve for r and n? If the energy (2.28 eV) equals Ke, why can't it knock the electron away, like with the photoelectric effect?

You can compare it to the photoelectric effect except that the equivalent of the work function is not the potential energy but the energy required to promote the electron from being in a bound state (that means $n$ is a finite integer) to the continuum (that means that $n$ is infinite). In other words, the energy price you have to pay in order to liberate the captive electron. As @etotheipi clarified, it is the total energy of the system not just the potential energy.

 

[kuruman]

I try to remind that the energy $E_n = \frac{-13.6 \text{eV}}{n^2}$ is the sum of a potential and a kinetic term, since from what I could tell OP had tried to "re-derive it" but had considered only the potential energy of interaction.

Yes, I see now what you were trying to do. Thanks for clarifying. I responded to OP accordingly.

 

[jjson775]

You can compare it to the photoelectric effect except that the equivalent of the work function is not the potential energy but the energy required to promote the electron from being in a bound state (that means $n$ is a finite integer) to the continuum (that means that $n$ is infinite). In other words, the energy price you have to pay in order to liberate the captive electron. As @etotheipi clarified, it is the total energy of the system not just the potential energy.

This is the answer to my question. As I said in the OP, I had already figured out you have to consider the total energy of the atom. The concept of the “bound state” is new to me. In the book I am studying, all I can find is a reference to Ionization energy of the Bohr atom.