Question Statement
According to the Geiger-Nuttall law for alpha decay, how does the half-life $T_{1/2}$ of an alpha-emitting radioactive nucleus relate to the kinetic energy $E_\alpha$ of the emitted alpha particle?
(A)
$\ln T_{1/2} = A - B \frac{Z}{\sqrt{E_\alpha}}$
(B)
$\ln T_{1/2} = A + B E_\alpha^2$
(C)
$\ln T_{1/2} = A - B \ln E_\alpha$
(D)
$\ln T_{1/2} = A + B \frac{\sqrt{E_\alpha}}{Z}$
Step-by-Step KaTeX Derivation & Detailed Explanation
### Step-by-Step Derivation & Verification: 1. **Gamow Barrier Penetration**: - An $\alpha$-particle of energy $E_\alpha$ trapped in a nuclear well of radius $R$ faces a repulsive Coulomb potential $V(r) = \frac{2 Z' e^2}{4\pi\varepsilon_0 r}$ out to turning point $b = \frac{2 Z' e^2}{4\pi\varepsilon_0 E_\alpha}$. 2. **WKB Tunneling Exponent ($2G$)**: $G = \frac{1}{\hbar} \int_R^b \sqrt{2m_\alpha (V(r) - E_\alpha)} dr \approx \frac{\pi Z' e^2}{\hbar \varepsilon_0} \sqrt{\frac{m_\alpha}{2 E_\alpha}} = K \frac{Z}{\sqrt{E_\alpha}}$ 3. **Decay Constant & Half-Life**: $\lambda = f \cdot e^{-2G} \implies \ln \lambda = \ln f - 2K \frac{Z}{\sqrt{E_\alpha}}$ $\ln T_{1/2} = \ln(\ln 2) - \ln \lambda = A - B \frac{Z}{\sqrt{E_\alpha}}$ 4. **Conclusion**: Option (A) is the authentic formulation of the Geiger-Nuttall law.