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The intense laser-atom physics and ionized electron

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xylai
#1
Dec10-08, 02:51 AM
P: 60
In the intense laser-atom physics, atom can eject high energy electron.
Corkumís 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 donít 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!
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olgranpappy
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Dec10-08, 05:32 PM
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what is the period of the laser?
xylai
#3
Dec10-08, 07:25 PM
P: 60
Quote Quote by olgranpappy View Post
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)

olgranpappy
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Dec11-08, 07:42 PM
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The intense laser-atom physics and ionized electron

and what is the velocity of the "high energy electron"?
xylai
#5
Dec11-08, 07:53 PM
P: 60
Quote Quote by olgranpappy View Post
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?
olgranpappy
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Dec12-08, 03:45 PM
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Quote Quote by xylai View Post
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
[tex]
v^2/2 \sim 0.1 {\rm a.u.}\;,
[/tex]
and the velocity of the electron is
[tex]
v\sim0.4 {\rm a.u.}\;.
[/tex]

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

Maybe this is all that the Corkum paper means by their statement.
xylai
#7
Dec12-08, 08:37 PM
P: 60
Quote Quote by olgranpappy View Post
So, assuming that a.u. means "Hartree atomic units" then the energy of the electron is
[tex]
v^2/2 \sim 0.1 {\rm a.u.}\;,
[/tex]
and the velocity of the electron is
[tex]
v\sim0.4 {\rm a.u.}\;.
[/tex]

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

Maybe this is all that the Corkum paper means by their statement.
Maybe your method is right.
But the period of the laser is 2*pi/w=31.4a.u. So I need consider this problem deeply.
Thank you for your kindly reply.
olgranpappy
#8
Dec13-08, 05:38 PM
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Quote Quote by xylai View Post
Maybe your method is right.
But the period of the laser is 2*pi/w=31.4a.u.
So I need consider this problem deeply.
You could consider that the "size" of the atom was never well defined and may as well be taken as 10 or 20 atomic units.

Thank you for your kindly reply.
You're welcome. Good luck solving your problem. Cheers.


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