Answer
$0.112$
Work Step by Step
Let $k$ and $k_{b}$ are the angular wave number of the wave functions in region $1$ and region $2$ respectively.
The boundary conditions for a particle in the give potential step give the following two equations:
$A+B=C$ ......................$(1)$
$k(A-B)=k_{b}C$
or, $A-B=\frac{k_{b}}{k}C$ ......................$(2)$,
In which $A$, $B$ and $C$ are constants.
solving equations $(1)$ and $(2)$, we get
$A=\frac{C}{2}(1+\frac{k_{b}}{k})$
and $B=\frac{C}{2}(1-\frac{k_{b}}{k})$
Therefore the reflection coefficient $(R)$ is
$R=\frac{|B|^{2}}{|A|^{2}}$
or, $R=\frac{|(1-\frac{k_{b}}{k})|^{2}}{|(1+\frac{k_{b}}{k})|^{2}}$
or, $R=\frac{|(k-k_{b})|^{2}}{|(k+k_{b})|^{2}}$
or, $R=\frac{|(1.45\times 10^{11}-7.23\times 10^{10})|^{2}}{|(1.45\times 10^{11}+7.23\times 10^{10})|^{2}}$
or, $R\approx0.112$
$\therefore$ The reflection coefficient is $0.112$.
