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So I was taught that:

Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy

If E(photon) = IE, then KE=0 of the electron.

What does this physically/theoretically mean?

My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?

Any clarifications would be greatly appreciated, thanks in advance.

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  • $\begingroup$ It simply means that it's speed after ionization will be zero, anyway you should be aware of that this is idealization that the electron left the potential well of the atom, thus in quantum physics this little bit different.. $\endgroup$
    – TMS
    Commented Dec 2, 2012 at 21:11
  • $\begingroup$ hmm, so is it like the speed of the electron decreases as it is pulled away from the atom..? Because it wouldn't make much sense for it to just stop after being ionized, right? $\endgroup$
    – student
    Commented Dec 2, 2012 at 22:04
  • $\begingroup$ Why not? imagine you are rolling a ball "across" a road of sand, if you give it a very big speed at start it will decrease due to friction with sand until it reaches the other side, and then will continue to move at constant speed (supposing that out of the road is very smooth), now if you give it less energy but at the same time enough to reach the other side of road, it's speed will continuously decrease but it will stop once reaches the edge. $\endgroup$
    – TMS
    Commented Dec 2, 2012 at 22:12
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    $\begingroup$ Well, atoms are always continuously moving with some speed, because they are never at 0K temperature. They are always moving/vibrating with some KE /viceversa. But so in the case of an electron, if it just 'stopped', is it at absolute zero temperature..? Doesn't it have SOME KE/movement? $\endgroup$
    – student
    Commented Dec 2, 2012 at 22:17
  • $\begingroup$ Yes you absolutely right, this why I said from the beginning that this an idealization, because this thermal vibration of atoms/electrons is usually negligibly small comparing to ionization energy. $\endgroup$
    – TMS
    Commented Dec 2, 2012 at 22:21

2 Answers 2

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This is the analogue of a projectile getting launched at exactly the escape velocity, something you may remember from studying gravity in freshman physics.

Here we're talking about the photoelectric effect. The electron jumps out of the material into air or vacuum, overcoming the force of attraction that tries to keep it bound inside the material (the force of attraction comes from the surface field, the image force effect, etc.).

If the electron jumps out with too little energy, it cannot escape, but gets pulled right back in. On the opposite extreme, if the electron jumps out with much more than enough energy to escape, it will not only break free of the material but also still energy left over, i.e. it will travel away from the material with a large kinetic energy.

You are asking about the borderline case. Here the electron has just barely enough energy to escape the material, with no energy left over. So it will slow down as it moves away from the material, and get slowed down more and more as it gets farther and farther away. It will never quite come to a stop, but its velocity will approach zero.

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  • $\begingroup$ Hi, Steve, I am learning about the same subject. I might not be getting your answer correctly. But if what you are saying is that when KE in that formula is equal to zero, the electron will still be traveling at escape velocity, which I guess is enough to ensure that the electron doesn't get pulled back into the atom, IT STILL HAS SOME VELOCITY AND THEREFORE K.E.. why is it not shown in the formula? $\endgroup$ Commented Aug 7, 2015 at 14:09
  • $\begingroup$ Just read over this, think the escape velocity makes sense when you look re hash the equation as varying in either time or distance from the gravitational body. $\endgroup$ Commented Feb 12, 2021 at 17:53
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Just read over this, think the escape velocity makes sense when you look re hash the equation as varying in either time or distance from the gravitational body and think of a single Hydrogen atom with one proton and one electron.

KE(t) + PE(t) = 0 KE(R) + PE(R) = 0

For a constant electron mass m, constant proton mass M, variable distance R, and distance varying velocity v(R). KE(R) + PR(R) = (1/2)mv(R)^2 -GmM/R = (1/2)v(R)^2 - GM/R The escape velocity would be the velocity that satisfies the above condition net zero energy condition.

So, v_esc(R) = sqrt(2GM/R)

If we consider the situation where the proton and electron theoretically occupy the same space (R approaches 0) we get.

v_esc(R=0) = inf

Which would mean that no amount of energy would ever separate them.

But another solution exists, when evaluated from some finite offset from the origin of the "gravitational" mass.

KE(R_i) + PE(R_i) = KE(R_f) + PE(R_f) (1/2)(v(R_i)^2 - v(R_f)^2) = -GM/(R_f - R_i) v(R_i)^2 - v(R_f)^2 = 2*M/(R_i - R_f)

Now we assume some constant fixed offset greater than zero (surface of proton/earth is different than center of proton/earth), and we want to know what the required initial velocity is to achieve a final velocity of zero as time or displacement approach infinity.

v(R_i)^2 - 0 = 2M/R_i - (2M/R_f -> inf) = v(R_i)^2 = 2M/R_i v_esc(R_i) = sqrt(2M/R_i)

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    $\begingroup$ Hi, welcome to Physics SE! Please see how to format equations using MathJax. $\endgroup$ Commented Feb 13, 2021 at 3:05

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