Hello.
I'm now calculating charge of the e-bunch generated from the photocathode which is hit by the laser pulse. The following is my condition.
1. Laser energy: 600 mJ.
2. Laser diameter: 7 mm.
3. Laser wavelength: 266 nm or 4.66 eV in photon energy (Forth harmonic of ND:YAG laser)
3. Q.E (Quantum Efficiency): 1e-5 (for copper)
In addition, laser pulse duration is 7 ns.
I simply thought the number of electrons in the bunch is the photon number within laser pulse times Q.E, thus the charge of the bunch is calculated as ~1e-5 C.
It seems too big.
e-bunch is supposed to be accelerated within 25 kV D.C voltage thus total current and average current density (Gaussian beam is assumed) are ~ 74 A and 3.43e+4 A/cm^(-2).
Is my calculation right? These values seem unpractically big as at least I know that typical e-bunch charge in most of the synchrotron is in order of sub nC.
Please clarify my confusion.
Generally, for a photocathode gun, for example, Cu cathode, the laser pulse energy should be seveal hundred micro-jourel. In your case, 600mJ is almost 3 order larger. I think the cathode would be destoryed by such a strong laser.
Thanks for answering my question.
But i though micro joule energy is for the
It is worthwhile to check the geometry of the gun with respect to space charge limitation (see A.Forrester, Large ion beams, Fundementals of generation and propagetion). In the case of the space charge limitation, first ejected electrons push the other electrons back in the cathode.
Thanks Pavel! You helped me to clarify my confusion. Our gun is just simple as plane copper. Do you have any idea about the space charge limitation for this case?
The critical parameter is the anode-cathode spacing a. For the case of the plain diode geometry Child equation gives the value of the space charge saturated current density as j=2.3*10-6 V1.5/a2 where V - applied voltage in Volts, a - anode-cathode spacing in cm, j - current density in A/cm2.
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