Answer
$ 4.84\;mA$
Work Step by Step
The reflection coefficient $R$ is given by
$R=\frac{|B|^2}{|A|^2}=\frac{|k-k_b|^2}{|k+k_b|^2}$
where, $k=\frac{2\pi\sqrt {2mE}}{h}$ and $k=\frac{2\pi\sqrt {2m(E-U_b)}}{h}$
The transmission coefficient (the probability of transmission) is
$T=1-R$
Putting the giving values,
$k=\frac{2\times\pi\sqrt {2\times 9.1\times 10^{-31}\times \frac{1}{2}\times 9.1\times 10^{-31}\times 900^2}}{6.63\times 10^{-34}}\;m^{-1}=7.46\times 10^{6}\;m^{-1}$
and
$k_b=\frac{2\times\pi\sqrt {2\times 9.1\times 10^{-31}\times (\frac{1}{2}\times 9.1\times 10^{-31}\times 900^2-1.6\times 10^{-19}\times 1.25\times 10^{-6})}}{6.63\times 10^{-34}}\;m^{-1}=5.25\times 10^{6}\;m^{-1}$
Therefore, $R=0.032$
$T=1-R=1-0.032=0.968$
Incident beam current is, $I_0=5\;mA$. Therefore, the transmitted current is given by
$I_T=TI_0$
$\implies I_T=0.968\times5\;mA$
$\implies I_T=4.84\;mA$
Therefore, the current on the other side of the step boundary is $ 4.84\;mA$
