Answer
a. No.
b. 0.488 nm.
Work Step by Step
a. A 0.100 kV potential difference gives the electron 0.100 keV of kinetic energy, which is much less than its rest energy of 0.511 MeV. We may use nonrelativistic expressions.
b. Find the de Broglie wavelength.
For a nonrelativistic particle, $p=mv$ and $KE=\frac{1}{2}mv^2$ so $p=\sqrt{2m(KE)}$.
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$
$\lambda=\frac{6.626\times10^{-34}J \cdot s}{\sqrt{2(9.11\times10^{-31}kg)(100eV)(1.60\times10^{-19}J/eV)}}$
$\lambda= 1.23\times10^{-10}m$
The electrons behave like waves and are diffracted by the slit.
For the first single-slit dark fringe, we have $a\;sin\theta=\lambda$. Solve for the slit width.
$$a=\frac{\lambda}{sin\theta}=\frac{1.23\times10^{-10}m }{sin\;14.6^{\circ}}=4.88\times10^{-10}m$$
The slit is on the order of an atomic spacing.
