Discussion
The radioactive \( {}^{210}_{84}\text{Po} \) emits alpha particles through the following decay process :
\[ \begin{aligned} {}^{210}_{84}\text{Po} &\rightarrow {}^{206}_{82}\text{Pb} \\ &+ {}^{4}_{2}\text{He} \end{aligned} \]
(Alpha Particle)
The height of the potential barrier experienced by the alpha particle emitted from radioactive nuclei \( {}^{210}_{84}\text{Po} \) is equal to :
Answer: A
The barrier height is about 26 MeV.
The barrier is the Coulomb energy of the alpha particle (charge 2e) touching the daughter nucleus (charge 82e). With \( e^{2}/4\pi\epsilon_{0} = 1.44 \) MeV fm and \( r_{0} \approx 1.2 \) fm:
\[ \begin{aligned} R &= 1.2\,(206^{1/3} + 4^{1/3}) \approx 9.0\ \text{fm} \\ V &= \frac{1.44 \times 2 \times 82}{9.0} \approx 26\ \text{MeV} \end{aligned} \]
The alpha particles of Po-210 have only about 5·3 MeV, far below this, which is why alpha decay proceeds by quantum tunnelling. 40, 42 and 80 MeV would need a much smaller contact radius.
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