Quanutm Mechanics and the conservation of energy?

I am looking at the basics of Quantum Mechanics and am trying to understand part of the issue regarding Einstein's equation for calculating Ek (Kinetic Energy).

Ek = hf - (work function) = h (f-fx)
*fx is the threshhold frequency - sorry, can't do subscripts

I understand that it takes the amount of energy fx in order to dislodge the electron.
This equation, however, seems to lose energy. What happens to the fx energy if the Ek is hf - hfx? Or perhaps the better question is: how is that E tracked/accounted for? It is my understanding that when said metal "recovers" the dislodged electron, fx amount of energy is released (as light?), correct?

2014-01-16T17:55:58Z

I understand the math given below, but I am still wondering about what happens to the energy which causes the electron to depart in the first place? How is that energy tracked/accounted for if not as part of the kinetic energy?

Also: threshold frequency = work function = f0 ... I was using fx instead of f0 since I can't do a subscript or the symbol for work function.

morningstar2014-01-16T18:02:56Z

Favorite Answer

The electron when attached to the atom has potential energy which is negative. PE = kQq/r; Q is positive and q is negative. It also has some kinetic energy. The magnitude of the potential energy is greater and its total energy is negative. The electron begins with negative energy, you add some energy, and it ends with positive energy, however not the full amount that you added.

Looking at it another way, if you return the electron to the atom, a photon will be emitted. Even if you bring the electron to a stop and then return it, a photon of frequency fx is emitted. If the electron had the full kinetic energy that you applied, and later could produce a photon, then energy would have been created, not conserved.

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elifino2014-01-17T03:38:26Z

Potential energy.

You know how if you were to throw a small object at extremely high speed (faster than the escape velocity) from the surface of the earth, its kinetic energy while it's near the ground would be less than the kinetic energy when it's millions of miles from earth? Same thing. Except the force is electrostatic instead of gravitational. The material attracts electrons. When an electron falls in from outside, that same potential energy is released, most likely as heat.

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?2014-01-17T01:07:49Z

The electron is knocked out and given kinetic energy.

E.g.

Photon energy = 7eV (this is what you refer to as 'hf')

Work function, Φ = 5eV (this is what you refer to as 'h.fx').
It is the energy needed to remove the electron from the metal.

There are therefore 2eV 'left over' after knocking out the electron. This is given to the electron as kinetic energy, (Ek) so the electron comes out at some speed which corresponds to it having kinetic energy Ek = 2eV.

So energy is conserved (Ek = hf - Φ = 7-5 = 2).

In fact it's a bit more complicated than that because electrons knocked out from just below the surface can lose some of their kinetic energy by collisions on the way out. So some of the electrons would have kinetic energy less than 2eV in this example.

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