Answer
$15\%$
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 $b=\sqrt {\frac{8\pi^2m(U-E)}{h^2}}$ ..............................$(2)$
Given, $E=5.1$ $eV$, $U=U_{b}=6.8$ $eV$
Substituting the above values in equation $2$, we get
$b=\sqrt {\frac{8\times\pi^2\times 9.1\times 10^{-31}\times(6.8-5.1)\times1.6\times 10^{-19}}{(6.63\times 10^{-34})^2}}$
or, $b\approx 6.668\times 10^{9}$ $m^{-1}$
As the change in kinetic energy is very small, we can write
$\Delta T=\frac{dT}{dE}\Delta E$ ..............................$(3)$
Now, differentiating both side of eq, $(1)$ with respect to $E$, we get
$\frac{dT}{dE}=-2Le^{-2bL}\frac{db}{dE}$
or, $\frac{dT}{dE}=-2LT\frac{db}{dE}$ ..............................$(4)$
Again, differentiating both side of eq, $(2)$ with respect to $E$, we get
$\frac{db}{dE}=\sqrt {\frac{8\pi^2m}{h^2}}\frac{d}{dE}\sqrt {(U-E)}$
$\frac{db}{dE}=-\sqrt {\frac{8\pi^2m}{h^2}}\frac{1}{2\times \sqrt {(U-E)}}$
or, $\frac{db}{dE}=-\sqrt {\frac{8\pi^2m(U-E)}{h^2}}.\frac{1}{2(U-E)}$
or, $\frac{db}{dE}=-\frac{b}{2(U-E)}$
Thus, eq. $(4)$ becomes,
$\frac{dT}{dE}=2LT\frac{b}{2(U-E)}$
And now eq. $(3)$ becomes,
$\Delta T=2LT\frac{b}{2(U-E)}\Delta E$
or, $\frac{\Delta T}{T}=L\frac{b}{(U-E)}\Delta E$ ..............................$(5)$
Given, $L=750$ $pm$ $=750\times 10^{-12}$ $m$
$\Delta E=1\%$ of $E =0.01\times 5.1$ $eV$ $=0.051$ $eV$
Therefore,
$\frac{\Delta T}{T}=750\times 10^{-12}\times\frac{6.668\times 10^{9}}{(6.8-5.1)}\times0.051$
or, $\frac{\Delta T}{T}=0.15$
or, $\frac{\Delta T}{T}=15\%$
Therefore, the percentage of change in the transmission coefficient is $15\%$.
