Difference between revisions of "Calculation of radiation yield"

From New IAC Wiki
Jump to navigation Jump to search
Line 39: Line 39:
 
<math>\gamma > 15</math>, <math>k<k_x</math>:
 
<math>\gamma > 15</math>, <math>k<k_x</math>:
  
<math>\Phi_n(Z,E_0,k) = \frac{p}{p_0}(\frac{4}{3}-2EE_0(\frac{p^2+p^2_0}{p^2p^2_0})+\frac{\omega_0E}{p^3_0}+\frac{\omega E_0}{p^3}-\frac{\omega\omega_0}{pp_0}+l{ \frac{k}{2pp_0} })</math>
+
<math>\Phi_n(Z,E_0,k) = \frac{p}{p_0}(\frac{4}{3}-2EE_0(\frac{p^2+p^2_0}{p^2p^2_0})+\frac{\omega_0E}{p^3_0}+\frac{\omega E_0}{p^3}-\frac{\omega\omega_0}{pp_0}+l[\frac{k}{2pp_0}(\omega_0(\frac{EE_0+p^2_0}{p^3_0})-)])</math>
  
  

Revision as of 21:18, 9 May 2008

The number of photons per MeV per incident electron per [math]g/cm^2[/math] of radiator (Z,A) is given by [*]:

[math]\frac{d^2n}{d\kappa dt} = \frac{3.495 \times 10^{-4}}{A\kappa}[Z^2\Phi_n(Z,E_0,k)+Z\Phi_e(Z,E_0,k)](MeV^{-1}g^{-1}cm^2)[/math],

where [math]\kappa[/math] - photon kinetic energy in MeV;

[math]E_0[/math] - incident electron total energy (in units of the electron rest mass);

[math]k[/math] - incident photon energy (in units of the electron rest mass);


Calculation of [math]\Phi_e(Z,E_0,k)[/math]

[math]\Phi_e(Z,E_0,k) = C_B\{2[1-\frac{2E}{3E_0}+(\frac{E}{E})^2][L-\sqrt{\eta}]+\sqrt{\eta}[1-\frac{L^2}{2\rho}-\frac{1}{\rho^2}(\frac{1}{2}L-[\frac{\rho(\rho+2)(E_0+1)}{E_0-1}]^{\frac{1}{2}})^2]\}[/math];

[math]E = E_0 - k[/math];

[math]\rho = E_0 -k(1+E_0-\sqrt{E^2 - 1})[/math];

[math]\eta = \rho/(\rho+2)[/math];

[math]L = 2 ln(\frac{(E_0-1)^{\frac{1}{2}}+[\eta(E_0+1)^{\frac{1}{2}}]}{(E_0-1)^{\frac{1}{2}}-[\eta(E_0+1)^{\frac{1}{2}}]})[/math];

[math]C_B = \frac{\frac{1}{4}\psi(\varepsilon)-1-lnZ^{\frac{2}{3}}}{3.798-ln\varepsilon-lnZ^{\frac{2}{3}}}[/math];

[math]\varepsilon = 100k/E_0EZ^{2/3}[/math];

Case A: For [math]\varepsilon \geq 0.88[/math] the screening effect is negligible, [math]\psi(\varepsilon)=19.19-4ln\varepsilon[/math] (free electron form) and in this case [math]C_B = 1[/math].

Case B: For [math]\varepsilon \lt 0.88[/math] we have [math]\psi(\varepsilon) = 19.70 + 4.117(0.88-\varepsilon)-3.806(0.88-\varepsilon)^2 + 31.84(0.88-\varepsilon)^3-58.63(0.88-\varepsilon)^4+40.77(0.88-\varepsilon)^5[/math]


Calculation of [math]\Phi_n(Z,E_0,k)[/math]

[math]\gamma(=100k/E_0EZ^{1/3}) \leq 15[/math] , [math]k\lt k_x[/math]:

[math]\Phi_n(Z,E_0,k) = 4([1+ (\frac{E}{E_0})^2][\frac{1}{4}\phi_1(\gamma)-\frac{1}{3}lnZ - f(Z)]-\frac{2E}{3E_0}[\frac{1}{4}\phi_2(\gamma)-\frac{1}{3}lnZ-f(Z)]) [/math]

[math]\gamma \gt 15[/math], [math]k\lt k_x[/math]:

[math]\Phi_n(Z,E_0,k) = \frac{p}{p_0}(\frac{4}{3}-2EE_0(\frac{p^2+p^2_0}{p^2p^2_0})+\frac{\omega_0E}{p^3_0}+\frac{\omega E_0}{p^3}-\frac{\omega\omega_0}{pp_0}+l[\frac{k}{2pp_0}(\omega_0(\frac{EE_0+p^2_0}{p^3_0})-)])[/math]




Reference: [*] J.L. Matthews, R.O. Owens, Accurate Formulae For the Calculation of High Energy Electron Bremsstrahlung Spectra, NIM III (1973) I57-I68.