## the intense laser-atom physics and ionized electron

In the intense laser-atom physics, atom can eject high energy electron.
Corkum’s work (PRL 71 1994 (1993)) says that the electron is not ionized immediately. Rather it stays in the vicinity of the ion for one or more laser periods.
I don’t know why the electron stays there for a while and why it is not ionized directly? Can you show me other reports about it?
Best wishes!
 Recognitions: Homework Help what is the period of the laser?

 Quote by olgranpappy what is the period of the laser?
The paper does not give the parameters of laser. The parameters I used are E0=0.1 a.u. and w=0.2 a.u. (E0: the amplitude; and w: angular frequency)

Recognitions:
Homework Help

## the intense laser-atom physics and ionized electron

and what is the velocity of the "high energy electron"?

 Quote by olgranpappy and what is the velocity of the "high energy electron"?
The energy of the the "high energy electron" is 0.1a.u., or 0.3 a.u., or 0.5 a.u.,......
But most of the electron have the energy 0.1a.u.
I don't know why you need know these values and how they help us solve the problem?

Recognitions:
Homework Help
 Quote by xylai The energy of the the "high energy electron" is 0.1a.u., or 0.3 a.u., or 0.5 a.u.,...... But most of the electron have the energy 0.1a.u. I don't know why you need know these values and how they help us solve the problem?
So, assuming that a.u. means "Hartree atomic units" then the energy of the electron is
$$v^2/2 \sim 0.1 {\rm a.u.}\;,$$
and the velocity of the electron is
$$v\sim0.4 {\rm a.u.}\;.$$

Then, since the size of an atom is about
$$3 {\rm a.u.}\;,$$
the time it takes the electron to "leave the atom" is about
$$3/0.4\sim 7.5 {\rm a.u.}\;.$$
This is on the order of the period of the laser. Right?

Maybe this is all that the Corkum paper means by their statement.

 Quote by olgranpappy So, assuming that a.u. means "Hartree atomic units" then the energy of the electron is $$v^2/2 \sim 0.1 {\rm a.u.}\;,$$ and the velocity of the electron is $$v\sim0.4 {\rm a.u.}\;.$$ Then, since the size of an atom is about $$3 {\rm a.u.}\;,$$ the time it takes the electron to "leave the atom" is about $$3/0.4\sim 7.5 {\rm a.u.}\;.$$ This is on the order of the period of the laser. Right? Maybe this is all that the Corkum paper means by their statement.
But the period of the laser is 2*pi/w=31.4a.u. So I need consider this problem deeply.