Answer
$-10\%$
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 potential height is very small, we can write
$\Delta T=\frac{dT}{dL}\Delta L$ ..............................$(3)$
Now, differentiating both side of eq, $(1)$ with respect to $L$, we get
$\frac{dT}{dL}=-2be^{-2bL}$
or, $\frac{dT}{dL}=-2bT$ ..............................$(4)$
And now eq. $(3)$ becomes,
$\Delta T=-2bT\Delta L$
$\frac{\Delta T}{T}=-2b\Delta L$
Given, $L=750$ $pm$ $=750\times 10^{-12}$ $m$
$\Delta L=1\%$ of $L$ $=0.01\times 750\times 10^{-12}$ $m$ $=750\times 10^{-14}$ $m$
Thus,
$\frac{\Delta T}{T}=-2\times 6.668\times 10^{9}\times 750\times 10^{-14}$
or, $\frac{\Delta T}{T}=-0.10$
or, $\frac{\Delta T}{T}=-10\%$
Therefore, the percentage of change in the transmission coefficient is $-10\%$.
