Answer
$5.071$ $eV$
Work Step by Step
The probability that a given particle of mass $m$ and energy $E$ will tunnel through a barrier of height $U_{b}$ and thickness $L$ is
given by the transmission coefficient $T$
$T\approx e^{-2bL}$, ..............................$(1)$
where $b=\sqrt {\frac{8\pi^2m(U_{b}-E)}{h^2}}$ ..............................$(2)$
Given, $T=0.0010$, $L=0.70$ $nm$$=7\times 10^{-10}$ $m$ and $U_{b}=6.0$ $eV$
Simplifying the equation $1$ and substituting the above values, we get
$-2bL=\ln{T}$
or, $b=\frac{1}{2L}\ln{\frac{1}{T}}$
or, $b=\frac{1}{2\times 7\times 10^{-10}}\ln{\frac{1}{0.0010}}$ $m^{-1}$
or, $b\approx 4.93\times 10^{9}$ $m^{-1}$
Now simplifying equation $2$ and substituting the above values, we get
$8\pi^2m(U_{b}-E)=b^2h^2$
or, $(U_{b}-E)=\frac{bh^2}{8\pi^2m}$
or, $(U_{b}-E)=\frac{(4.93\times 10^{9})^2\times (6.63\times 10^{-34})^2}{8\times\pi^2\times 9.1\times 10^{-31}}$ $J$
or, $(U_{b}-E)\approx 1.487\times 10^{-19}$ $J$
or, $(U_{b}-E)=0.929$ $eV$
$\therefore E=U_{b}-0.929$ $eV$
or, $E=6.0-0.929$ $eV$
or, $E=5.071$ $eV$
Therefore, the energy of an incident electron is $5.071$ $eV$
