Answer
$8.33\times 10^{103}$ $years$
Work Step by Step
The number of protons striking the potential energy barrier per second is
$n=\frac{I}{q}$ ........................$(1)$,
where $q$ is the charge of proton.
Substituting the given value $I=1000$ $A$ in eq. $(1)$, we get
$n=\frac{1000}{1.6\times 10^{-19}}$ $s^{-1}$
or, $n=6.25\times 10^{21}$ $s^{-1}$
Let, $N_{I}$ number of protons strikes the potential barrier in time $t$.
Thus, $N_{I}=nt$ ........................$(2)$
If $T$ be the transmission coefficient, then the number of protons transmitted is given by
$N_{T}=TN_{I}$
Substituting eq. $(2)$, we get
$N_{T}=Tnt$ ........................$(3)$
If we consider that $t_{1}$ is the waiting time for which the one proton will be transmitted. Now eq. $(3)$ becomes
$1=Tnt_{1}$
or, $t_{1}=\frac{1}{nT}$ ........................$(4)$
Now, 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 $T$
$T\approx e^{-2bL}$, ..............................$(5)$
where $b=\sqrt {\frac{8\pi^2m(U_{b}-E)}{h^2}}$ ..............................$(6)$
Given, $E=5.0$ $eV$, $U_{b}=6.0$ $eV$$L=0.70$ $nm$ $=7\times 10^{-10}$ $m$
Substituting the above values in equation $6$, we get
$b=\sqrt {\frac{8\times\pi^2\times 1.67\times 10^{-27}\times(6-5)\times1.6\times 10^{-19}}{(6.63\times 10^{-34})^2}}$
or, $b\approx 2.191\times 10^{11}$ $m^{-1}$
The (dimensionless) quantity $2bL$ is then
$2bL=2\times (2.191\times 10^{11})\times (7\times 10^{-10})\approx 306.74$
and, from the equation $1$, the transmission coefficient is
$T\approx e^{-2bL}= e^{-306.74}=6.0885\times 10^{-134}$
Thus, from eq. $(4)$, we get
$t_{1}=\frac{1}{6.25\times 10^{21}\times6.0885\times 10^{-134}}$ $s$
or,$t_{1}=2.628\times 10^{111}$ $s$
or, $t_{1}=8.33\times 10^{103}$ $years$
Therefore, we have to wait $8.33\times 10^{103}$ $years$ for one proton to be transmitted.
