Discussion
To penetrate the Coulomb barrier of a light nucleus (\({}^{1}_{1}H\)), a proton must have a minumum energy of the order of (Consider radius of nucleus ~ 1Fm) :
Answer: C
The proton needs an energy of the order of 1.44 MeV.
The Coulomb barrier is the electrostatic potential energy of the two charges when they touch, here at a separation of about 1 fm:
\[ \begin{aligned} U &= \frac{e^{2}}{4\pi\varepsilon_{0}r} = \frac{1.44\ \text{MeV fm}}{1\ \text{fm}} \\ &= 1.44\ \text{MeV} \end{aligned} \]
using the convenient constant \(e^{2}/4\pi\varepsilon_{0} = 1.44\) MeV fm. Nuclear barriers are of the order of MeV, which is why nuclear reactions need MeV particles; TeV and GeV are far too large, and keV far too small, although quantum tunnelling lets fusion in stars proceed at keV thermal energies.
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