Answer
$9.24\times 10^{-6}$
Work Step by Step
The probability that a proton 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, $E=3.0$ $MeV$, $U_{b}=10$ $MeV$$L=10$ $fm$ $=1\times 10^{-14}$ $m$
Substituting the above values in equation $2$, we get
$b=\sqrt {\frac{8\times\pi^2\times 1.67\times 10^{-27}\times(10-3)\times 10^6\times1.6\times 10^{-19}}{(6.63\times 10^{-34})^2}}$
or, $b\approx 5.796\times 10^{14}$ $m^{-1}$
The (dimensionless) quantity $2bL$ is then
$2bL=2\times (5.796\times 10^{14})\times (1\times 10^{-14})\approx11.592$
and, from the equation $1$, the transmission coefficient is
$T\approx e^{-2bL}= e^{-11.592}=9.24\times 10^{-6}$
Therefore, the transmission coefficient is $9.24\times 10^{-6}$
