= Solution
Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In <quantum mechanics> the <wavefunction> penetrates the classically forbidden side as an <evanescent wave>. Define the <tunnelling decay constant>
$$
\kappa=\frac{\sqrt{2m(V_0-E)}}\hbar.
$$
Boundedness at positive infinity excludes $e^{+\kappa x}$, so
$$
\psi(x)=\begin{cases}Ae^{ikx}+Be^{-ikx},&x\le0,\\Ce^{-\kappa x},&x>0.\end{cases}
$$
Matching the <wavefunction> and its <derivative> gives
$$
\frac BA=\frac{k-i\kappa}{k+i\kappa},\qquad\frac CA=\frac{2k}{k+i\kappa}.
$$
The reflected <probability amplitude> has unit <modulus>, hence <quantum reflection probability> one. The decaying tail carries zero <probability current>: this semi-infinite <potential step> has no transmitted beam at infinity, unlike a finite <potential barrier>. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex <wavefunction>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022-q9-step-penetration.png]
{title=Real and imaginary components of a matched <wavefunction> at a <potential step>, with <exponential decay> of the <probability density>}
Because the <probability density> decays as $|C|^2e^{-2\kappa x}$, the requested $1/e$ penetration distance is
$$
\boxed{d=\frac1{2\kappa}=\frac{\hbar}{2\sqrt{2m(V_0-E)}}.}
$$
For a <proton>, take $m_p=1.67262\times10^{-27}\,\mathrm{kg}$, $V_0-E=10\,\mathrm{MeV}=1.60218\times10^{-12}\,\mathrm J$ and $\hbar=1.05457\times10^{-34}\,\mathrm{J\,s}$. Then
$$
\boxed{d=7.20\times10^{-16}\,\mathrm m=0.720\,\mathrm{fm}.}
$$
The <wavefunction> amplitude decay length $1/\kappa=1.44\,\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\,\mathrm{MeV}$ is small compared with the <proton> <rest energy>, approximately $938\,\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the <Planck constant>, the displayed differential equation gives $d_{\rm literal}=2\pi d=4.53\,\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\,\mathrm{fm}$ is the physical estimate using the standard <Schrödinger equation>.
Back to article page
