Answer
$1.7883\;mA$
Work Step by Step
The probability that an electron of mass $m$ and energy $E$ will tunnel through a barrier of height $U$ and thickness $L$ is
given by the transmission coefficient $T$
$T\approx e^{-2bL}$, ..............................$(1)$
where $b=\sqrt {\frac{8\pi^2m(U-E)}{h^2}}$ ..............................$(2)$
The speed of each electron is $v=1.2\times 10^{3}\;m/s$. Therefore, the energy of each electron is given by
$E=\frac{1}{2}m_ev^2$
or, $E=\frac{1}{2}\times9.1\times 10^{-31}\times(1.2\times 10^{3})^2\;J$
or, $E=6.552\times 10^{-25}\;J$
or, $E=4.095\times 10^{-6}\;eV$
The beam of electrons encounters a potential barrier of height $-4.719\;\mu V$ and thickness $200.0\;nm$. Therefore, $U=4.719\times 10^{-6}\;eV$ and $L=200\;nm$
Substituting the above values in equation $2$, we get
$b=\sqrt {\frac{8\times\pi^2\times 9.1\times 10^{-31}\times(4.719-4.095)\times 10^{-6}\times1.6\times 10^{-19}}{(6.63\times 10^{-34})^2}}\;m^{-1}$
or, $b\approx 4.04\times 10^{6}$ $m^{-1}$
The (dimensionless) quantity $2bL$ is then
$2bL=2\times (4.04\times 10^{6})\times (200\times 10^{-9})=1.616$
and, from the equation $1$, the transmission coefficient is
$T\approx e^{-2bL}= e^{-1.616}\approx 0.1987$
Incident beam current is, $I_0=9\;mA$. Therefore, the transmitted current is given by
$I_T=TI_0$
$\implies I_T=0.1987\times9\;mA$
$\implies I_T=1.7883\;mA$
