= Solution
Set $k=\sqrt{2mE}/\hbar$ and $\kappa=\sqrt{2m(U-E)}/\hbar$. For unit incident <amplitude>, use $\psi=e^{ikx}+r e^{-ikx}$ on the left and $\psi=t_s e^{-\kappa x}$ on the right; the growing exponential is excluded at infinity. Continuity at zero gives $1+r=t_s$ and $ik(1-r)=-\kappa t_s$. Thus
$$
r=\frac{k-i\kappa}{k+i\kappa}=e^{i\delta},\qquad |r|^2=1,
\qquad\boxed{\delta=-2\arctan(\kappa/k)=-2\arccos\sqrt{E/U}.}
$$
The phase is chosen in $(-\pi,0)$ for $0<E<U$. The <evanescent wave> is $\psi(x)=2k\,e^{-\kappa x}/(k+i\kappa)$ for $x>0$. Its <probability density> is nonzero, but its <probability current> $j=(\hbar/m)\operatorname{Im}(\psi^*\psi')$ is zero. Penetration into a forbidden region therefore does not require transmitted flux through a semi-infinite step.
For a finite rectangular barrier, matching both interfaces gives
$$
\mathcal T=\left[1+\frac{U^2}{4E(U-E)}\sinh^2(\kappa a)\right]^{-1}
\simeq\frac{16E(U-E)}{U^2}e^{-2\kappa a}.
$$
The approximation requires \b[$\kappa a\gg1$], or a width much greater than the evanescent penetration <length>. The tunnelling exponent is $-2a\sqrt{2m(U-E)}/\hbar$; its factor of two converts an <amplitude> decay to a <probability> decay. For the <electron> parameters, $\kappa=1.984\times10^9\,\mathrm{m^{-1}}$ and $\kappa a=19.84$. Therefore
$$
\boxed{\mathcal T\simeq3e^{-39.68}=1.75\times10^{-17}.}
$$
Keeping only the exponential gives $5.83\times10^{-18}$; the matched-interface prefactor accounts for the factor of three.
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