Comptonov efekt (TEK U IZRADI! NE UPOTREBLJAVATI!)

Compton je 1922. primijenio fotonski koncept (pojam fotona, fotonsku sliku) elektromagnetskog zračenja da bi objasnio raspršenje X-zraka na elektronima. U njemu dolazi do promjene valne duljine odnosno frekvencija elektromagnetskog zračenja, što se ne moze objasniti klasičnom fizikom. Compton je pronašao da je razlika tih valnih duljina fotona dana s

$$\Delta \lambda = \lambda' - \lambda = \lambda_{c} (1 - \cos \theta)$$ (1.19)

$$\lambda_{c} = \frac {h}{mc} = 2,43 \cdot 10^{-10} cm$$ (1.20)

$$E_{\gamma} = h \nu = h \frac {c}{\lambda}$$ (1.21)

$$p_{x} = h \nu/c = h/\lambda$$ (1.22)

 

p_y = 0 (1.23)

$${\cal E}_{e} = m c^{2}$$ (1.24)

 

P_x = 0 (1.25)

 

P_y = 0 (1.26)

$$\gamma = \frac {1}{\sqrt {1 - \beta^{2}}}$$ (1.27)

$$\beta = v/c$$ (1.28)

$$E^{'}_{\gamma} = h \nu'$$ (1.29)

$$p'_{x} = (h/\lambda') \cos \theta$$ (1.30)

$$p_{y} = (h/\lambda) \sin \theta$$ (1.31)

$${\cal E}^{'}_{e}= \gamma_{m} c^{2}$$ (1.32)

$$P'_{x} = m \gamma \beta c \cos \phi$$ (1.33)

$$P'_{y} = m \gamma \beta c \sin \phi$$ (1.34)

$$\gamma = \frac {1}{\sqrt {1 - \beta^{2}}}$$ (1.35)

$$\beta = v/c$$ (1.36)

Evo objašnjenja preko sudara elektrona s česticom svjetlosti, fotonom:
Sačuvanje energije:

$$E_{\gamma} + {\cal E}_{e} = E'_{\gamma} + {\cal E}^{'}_{e}$$ (1.37)

$$h \nu + m c^{2} = h \nu' + m \gamma e^{2}/ : me^{2}$$ (1.38)

$$\frac {h}{mc} (\frac {1}{\lambda} - \frac {1}{\lambda'}) = \gamma - 1$$ (1.39)

Sačuvanje impulsa

p_x + 0 = p'_x + P'_x (1.40)

 

0+0 = p'_y + P'_y (1.41)

$$\frac {h}{\lambda} = m \gamma \beta c \cos \phi + \frac {h}{\chi} \cos \theta$$ (1.42)

$$0= - m \gamma_{\beta} c \sin \phi + \frac {h}{\lambda'} \sin \theta$$ (1.43)

$$(\frac {h}{\lambda} - \frac {h}{\lambda'} \cos \theta)^{2} = (m \gamma \beta c)^{2} \cos^{2} \phi$$ (1.44)

$$+ (\frac {h}{\lambda'})^{2} \sin^{2} \theta = (m \gamma \beta c)^{2} \sin^{2} \phi$$ (1.45)

$$h^{2} (\frac {1}{\lambda^{2}} - \frac {2}{\lambda \lambda'} \... ...^{'2}} \cos^{2} \theta) = (m c \gamma \beta)^{2} \cos^{2} \phi$$ (1.46)

$$h^{2} \frac {1}{\lambda^{'2}} sin^{2} \theta = (m c \gamma \beta)^{2} sin^{2} \phi$$ (1.47)

$$\longrightarrow h^{2} (\frac {1}{\lambda^{2}} - \frac {2}{\la... ...\frac {1}{\lambda^{'2}}) = (m c \gamma \beta)^{2}/ : (m c)^{2}$$ (1.48)

$$(\frac {h}{mc})^{2} [\frac {1}{\lambda^{2}} - \frac {2}{\lamb... ...2}{\lambda \lambda'} (1 - \cos \theta)] = \gamma^{2} \beta^{2}$$ (1.49)

Sačuvanje energije $\rightarrow$

$$\Longrightarrow \frac {1}{\lambda} - \frac {1}{\lambda'} = \f... ...ac {1}{\lambda \lambda'} (1 - \cos \theta) / \cdot \lambda^{'}$$ (1.50)

$$\Longrightarrow \Delta \lambda = \lambda' - \lambda = (\frac {h}{mc}) (1 - \cos \theta)$$ (1.51)

Ponovi isto na vektorskoj rotaciji! ( $\vec {p}_{\gamma} = \hbar \vec {k} , \vec {k} \cdot \vec {k'} = k k' \cos \theta$)

Sačuvanje impulsa u 3-vektorskom obliku:

$$\vec {p} + 0 = \vec {p'} + \vec {P'}$$ (1.52)

$$\vec {p} - \vec {p'} = \vec {P'} /^{2}$$ (1.53)

$$\vec {p}^{2} - 2 \vec {p} \cdot \vec {p'}^{2} = \vec {P'}^{2}... ... \gamma c \vec {\beta})^{2} = m^{2} c^{2} \gamma^{2} \beta^{2}$$ (1.54)

$$\vert\vec {p}\vert \vert\vec {p'}\vert \cos \theta = \vert\vec {p}\vert \vert \vec {p'}\vert [1-(1- \cos \theta)]$$ (1.55)

$$\vert\vec {p}\vert^{2} - 2 \vert\vec {p}\vert \vert\vec {p'}\... ...1 - \cos \theta) = (m c)^{2} \gamma^{2} \beta^{2}/ : (m c)^{2}$$ (1.56)

$$(\vert\vec {p}\vert - \vert\vec {p'}\vert)^{2} = h^{2} (\frac {1}{\lambda} - \frac {1}{\lambda'})^{2}$$ (1.57)

$$\frac {h}{\lambda} - \frac {h}{\lambda'}$$ (1.58)

$$\gamma^{2} - 1$$ (1.59)

$$(\gamma - 1)^{2} + 2 (\gamma-1)$$ (1.60)

$$\gamma - 1 = \frac {h}{m e} (\frac {1}{\lambda} - \frac {1}{\... ...ac {1}{\lambda} = \frac {1}{\lambda'})/ \cdot \lambda \lambda'$$ (1.61)

$$\Longrightarrow \lambda' - \lambda = (\frac {h}{mc} (1 - \cos \theta)$$ (1.62)

U 4-vektorskom obliku: p + P = p' + P'

$$\Longrightarrow p - p' = p' - p/^{2} \Longrightarrow p^{2} - 2 p \cdot p' + p'^{2} = P'^{2} - 2 P \cdot P' + P^{2}$$ (1.63)

$$\Longrightarrow - 2 p \cdot p' = 2 (m c^{2})^{2} - 2 P \cdot P' \Longrightarrow - p \cdot p' = (m c^{2})^{2} - P \cdot P'$$ (1.64)

$$\Longrightarrow - \frac {1}{c^{2}} E_{\gamma} E'_{\gamma} + \... ...{h^{2}}{\lambda \lambda'} \cos \theta = (m c)^{2} (\gamma - 1)$$ (1.65)

$$- \frac {1}{c^{2}} \frac {h c}{\lambda} \cdot \frac {hc}{\lambda'} \vert\vec {p}\vert \vert\vec {p'}\vert \cos \theta$$ (1.66)

$$\lambda \lambda' \cdot / h (1 - \cos \theta) = m c (\lambda' ... ...ghtarrow \lambda' - \lambda = \frac {h}{mc} (1- \cos \theta)$$ (1.67)

U 4-vektorskom obliku najlakše, ali smo i tu morali koristiti izraz $(\gamma - 1) = \frac {h}{mc} (\frac {1}{\lambda} - \frac {1}{\lambda'})$ koji smo dobili iz sačuvanja energije.