Fjod Vaje Teden07

Pri razpadu \( \beta^- \) se pretvarja nevtron v proton in številke se spremenijo na sledeč način na \( \beta^+ \).

\begin{align*} Z' &= Z - 1 \\ N' = N + 1 \end{align*}

Poglejmo ohranitev energij. Predpostavimo, da \( X \) na začetku pri miru

\begin{align*} \vec{p}_X &= \vec{0} \\ E_{start} &= m_{X} && c = 1 \end{align*}

Na koncu pa imamo enačbe

\begin{align} \vec{p}_Y + \vec{p}_e + \vec{p}_{\nu} &= \vec{0} \\ E_{end} &= E_Y + E_e + E_{\nu} = \\ &= m_Y + m_e + T_e + \cancel{m_{\nu}} + T_{\nu} \\ &= m_Y + m_e + T_e + T_{\nu} \end{align}

Tukaj smo zanemarili kinetično energijo jedra, because it is just so massive. (That’s what she said).

Izenačimo začetno in končno energijo in dobimo, da je sproščena energija \( Q \)

\begin{align*} E_{start} &= E_{end} \\ Q &= m_X - m_Y - m_E = T_e + T_{\nu} \end{align*}

Razpadno širino \( X \) definiramo kot

\[ \Gamma(X \to Y e^+ \nu) \]

Uporabimo Fermijevo zlato pravilo

\begin{equation} \label{eq:1} \mathrm{d} \Gamma = \left| \left\langle T_{fi} \right\rangle \right| ^2 \frac{\mathrm{d} ^3 \vec{p}_Y}{\left(2 \pi \right)^3} \frac{\mathrm{d} ^3 \vec{p}_e}{\left( 2\pi \right)^3} \frac{\mathrm{d} ^3 \vec{p}_{\nu}}{\left( 2\pi \right)^3} \delta^{(3)} \left( \vec{p}_Y + \vec{p}_e + \vec{p}_{\nu} \right) 2 \pi \delta \left( m_X - E_Y - E_e - E_{\nu} \right) \end{equation}

Naš fazni prostor ni Lorentzovo invarianten:

\[ \left\langle \vec{p}' | \vec{p} \right\rangle = \left( 2 \pi \right) ^3 \delta^{(3)} \left( \vec{p} - \vec{p}' \right) \]

Lorentzovo invarianten ima še en dodaten člen \( 2E \):

\[ \left\langle \vec{p} \, ', E' | \vec{p}, E \right\rangle = \left( 2 \pi \right) ^4 2 E \delta^{(3)} \left( \vec{p} - \vec{p}\, ' \right) \delta \left( E - E' \right) \]

Če nadaljujemo enačbo \ref{eq:1}, se nam

\[ \mathrm{d} ^3 \vec{p}_i = p_Y ^2 \mathrm{d} p_Y \mathrm{d} \Omega_Y; \, i = Y, e, \nu \]

Velja

\[ \int\limits_{}^{} \frac{p_Y ^2 \mathrm{d} p_Y \mathrm{d} \Omega}{\left( 2 \pi \right) ^3} \delta^{(3)} \left( \vec{p}_Y + \vec{p}_e + \vec{p}_{\nu}\right) = \frac{1}{\left( 2 \pi \right) ^3} \int\limits_{}^{} \mathrm{d}p_{Y_x} \mathrm{d} p_{Y_y} \mathrm{d} P_{Y_z} \delta^{(3)} \left( \vec{p}_Y + \vec{p}_e + \vec{p}_{\nu} \right) = 1 \]

Izrazimo

\begin{align*} -\vec{p}_Y &= \vec{p}_e + \vec{p}_{\nu} \\ E_y &= \sqrt{\left( \vec{p}_e + \vec{p}_{\nu} \right) ^2 + m_Y ^2} \end{align*}

Tako lahko nadalje zapišemo kot

\begin{align*} \int\limits_0^{4 \pi} \int\limits_0^{\infty} \mathrm{d} \Gamma &= \frac{\left| \left\langle T_{fi} \right\rangle \right| ^2}{\left( 2 \pi \right) ^5} \delta \left( m_X - \underbrace{\sqrt{\left( \vec{p}_e + \vec{p}_{\nu} \right) ^2 + m_Y ^2}}_{\sim m_Y} - \underbrace{\sqrt{m_e ^2 + \vec{p}_e ^2}}_{= m_e + T_e} - p_{\nu} \right) p_e ^2 \mathrm{d} p_e \mathrm{d} \Omega_e p_{\nu} ^2 \mathrm{d} p_{\nu} \mathrm{d} \Omega_{\nu} \left( 2 \pi \right) ^3 && T_{fi} = \mathrm{konst} \\ &= \frac{\left| \left\langle T_{fi} \right\rangle \right| ^2}{\left( 2 \pi \right) ^5} \delta \left( \underbrace{m_X - m_Y - m_e}_Q - T_e - T_{\nu} \right) p_e ^2 \mathrm{d} p_e \mathrm{d} \Omega_e p_{\nu} ^2 \mathrm{d} p_{\nu} \mathrm{d} \Omega_{\nu} \left( 2 \pi \right) ^3 \\ &= \frac{\left| \left\langle T_{fi} \right\rangle \right| ^2}{\left( 2 \pi \right) ^3} \delta \left( Q - T_e - T_{\nu} \right) \left( 4 \pi \right) ^2 p_e ^2 \mathrm{d} p_e p_{\nu} ^2 \mathrm{d} p_{\nu} \\ \int\limits_{p_{\nu}}^{} \mathrm{d} \Gamma &= 4 \left| \left\langle T_{fi} \right\rangle \right| ^2 \int\limits_{}^{} \delta \left( Q - T_e - T_{\nu} \right) p_{\nu} ^2 \mathrm{d} p_{\nu} p_e ^2 \mathrm{d} p_e \end{align*}

Za nevtrino smo rekli, da je \( T_{\nu} = p_{\nu} \), kar pomeni, da

\[ Q - T_e - T_{\nu} = Q - T_e - T_{\nu} \implies \ p_{\nu} = Q - T_e \]

Z upoštevanjem tega zgornji integral postane

\begin{align*} \int\limits_{p_{e}}^{} \mathrm{d} \Gamma &= 4 \left| \left\langle T_{fi} \right\rangle \right| ^2 \left( Q - T_e \right) ^2 p_e ^2 \mathrm{d} p_e \\ &= 4 \left| \left\langle T_{fi} \right\rangle \right| ^2 \left( Q -T_e \right) p_e ^2 \mathrm{d} p_e && m_e \sim 0 \to T_e = p_e \\ \int\limits_{p_e} \Gamma &= \int\limits_0^Q 4 \left| \left\langle T_{fi} \right\rangle \right| ^2 \left( Q -p_e \right) p_e ^2 \mathrm{d} p_e \\ \implies \Gamma &= \left| \left\langle T_{fi} \right\rangle \right| Q^5 \left( \frac{1}{3} - \frac{2}{4} + \frac{1}{5} \right) = \frac{\left| \left\langle T_{fi} \right\rangle \right| ^2 Q^5}{30} \end{align*}

\begin{equation} \label{eq:2} \mathrm{d} \Gamma = 4 T_{fi} ^2 \left( Q - p_e \right) ^2 p_e ^2 \mathrm{d} p_e \end{equation}

Vzamemo enačbo \ref{eq:2} in jo delimo z \( p_e ^2 \), potem dobimo

\begin{equation} \label{eq:3} K \left( p_e \right) \sqrt{\frac{1}{p_e ^2} \frac{\mathrm{d} \Gamma}{\mathrm{d} p_e}} = k \cdot \left( Q - p_e \right) \end{equation}

Dobimo t.i Kurie plot, kjer smo privzeli, da je \( m_{\mu} = 0 \), in če ni, se rep ukloni.

Vir slike: (??, ????)

Gledamo razpad \( \alpha \) \( U^{238} \), ki razpade na \( ^{234}Th \) in \( \alpha \), ki sta oba pozitivno nabita.

Radij jedra je

\[ R = R_{Th} + R_{\alpha} \]

Definiramo \( \bar{R} \), ki je točka, kjer Coulumbski potencial \( V_C \) postane enak sproščeni energiji \( Q \).

Sproščena energija je ista kot prej.

Valovna funkcija \( \alpha \) lahko tunelira in je lahko znotraj ali zunaj potencialne jame.

Zanima nas, kako hitro bomo pobegnili potencialni jami.

Z \( P_T \) označimo verjetnost za tuneliranje in z \( f \) frekvenco pred bariero, ki je definirana kot

\begin{equation} \label{eq:4} f = \frac{v_j}{R} \end{equation}

Zanima nas tudi verjetnost za \( P_T \) in zato si pogledamo verjetnost za tuneliranje skozi majhno bariero \( \mathrm{d} r \), Na levi strani bariere imamo \( \Psi_L \), na desni strani pa \( \Psi_D \).

Tranzitivnost zapišemo kot

\[ T = \frac{\left| \Psi_D \right| ^2}{\left| \Psi_L \right|^2} \]

Enako velja tudi, če razširimo

\begin{equation} \label{eq:5} T = \frac{\left| \Psi \left( r + \mathrm{d} r \right) \right| ^2}{\left| \Psi (r) \right| ^2} \end{equation}

Kaj bosta \( \Psi_L \) in \( \Psi_D \) v sferičnem sistemu?

Odgovor je

\[ \Psi = R_{nl} Y_{lm} \]

in predpostavimo \( l = 0, m = 0 \) in tako dobimo

\[ \Psi =R_{00} = \frac{e^{\iu kr }}{r} \]

v delu, kjer \( Q < V_C \).

\( k \in \Im \to \iu \kappa \left( r \right) \) iz česar sledi

\[ \Psi = \frac{e^{-\kappa \left( r \right) r}}{r} \simeq \frac{e^{-\kappa r}}{r} \]

Predpostavimo odvisnost \( \kappa \) od \( r \), saj je diferencial zelo majhen. Tako tranzitivnost \ref{eq:5} postane

\begin{align*} T &= \left| \frac{e^{- \kappa \left( r + \mathrm{d} r \right)}}{r + \mathrm{d} r} \right| \cdot \left| ^2 \frac{r}{e^{- \kappa r}} \right| ^2 \\ &= \frac{e^{-\kappa \left( r + \mathrm{d}r \right)}}{r ^2 + \cancel{2r \mathrm{d} r} +\cancel{ \mathrm{d} r ^2 }} \cdot \frac{r ^2}{e^{-2 \kappa r}} && \text{they be small} \\ &= e^{- 2 \kappa \mathrm{d} r} = \mathrm{d} P_T \end{align*}

Vrnimo se nazaj na verjetnost za prehod

\begin{align*} P_T &= \prod \mathrm{d} P_T \\ \ln P_T &= \ln \left( \prod_{i}^{} \mathrm{d}P_T \right) \\ &= \sum\limits_i^{} \ln \mathrm{d} P_T \\ &= \int\limits_R^{\bar{R}} \ln \mathrm{d} P_T \\ &= \int\limits_R^{\bar{R}} - 2 \kappa \mathrm{d} r = -2 \int\limits_R^{\bar{R}} \kappa \left( r \right) \, \mathrm{d} r \end{align*}

Razpoložljiva energija je tako

\begin{align*} Q &= T + V_c \\ Q &= \frac{\kappa ^2}{2 \mu} + V_c \\ \frac{\kappa ^2}{2 \mu} &= \frac{Z_{\alpha} Z' e_0 ^2}{r} - Q = \frac{\kappa ^2 \left( r \right)}{2\mu} \end{align*}

Nadaljujemo zgornji integral z uporabo nove spremenljivke \( y = \frac{r}{\bar{R}} \) in \( \bar{R} = \frac{Z_{\alpha} Z' e_0 ^2}{4 \pi R} \)

\begin{align*} - 2 \int\limits_R^{\bar{R}} \sqrt{2 \mu \left(\frac{Z_{\alpha} Z' e_0 ^2}{r} - Q\right)} \, \mathrm{dr} &= -2 Z_{\alpha} Z' e_0 ^2 \sqrt{\frac{2\mu}{Q}} \int\limits_{\frac{R}{\bar{R}}}^1 \sqrt{\frac{1}{y} - 1} \, \mathrm{d}y \\ &= \sqrt{y - y ^2} - \mathrm{arctg} \left( \sqrt{\frac{1}{y} - 1} \right) \end{align*}

Uvedemo novo količino, kjer je \( G \) Ganow faktor in ima vrednost

\[ G = - Z_{\alpha}Z' e_0 ^2 \sqrt{\frac{2\mu}{Q}} \left( \mathrm{arctg} \sqrt{\frac{\bar{R} - R}{R}} - \sqrt{\left( \frac{R}{\bar{R}} \right) - \left( \frac{R}{\bar{R}} \right)^2 } \right) \]

\[ \Gamma = \frac{v_{jama}}{R} e^{-2G} \]

kjer je \( v_{jama} \) hitrost delca v jami.

Pazi: Koliko je \( R \) v primerjavi z \( \overline{R} \)? Močno odvisna je od delca in npr. za \( ^{238}U \) so lastnosti:

\begin{align*} v_j &= 0.06 \\ R &\sim 9.3 \mathrm{fm} \\ \overline{R} & \sim 63 \mathrm{fm} \\ \implies G &= 45; \ \Gamma = 2 \cdot 10^{-39} \mathrm{MeV} \implies t = 7 \cdot 10^9 \mathrm{let} \end{align*}